. 5
( 14)


(n + 3n4 + 4n3 + 2n2 + 2n).

5.3.3 A Variant
We now consider a slightly di¬erent question. How many distinct colorings of the vertices
of a regular hexagon are there if we are forced to color exactly three vertices blue and
three vertices yellow? The group G is the same as before, but the set S is di¬erent. Of
the 64 possible colorings of the vertices, only ( 6 ) = 20 are legal, since 3 vertices out
of 6 are chosen to be colored blue; the other vertices must be colored yellow. If π is a
permutation of G, then within each cycle of π, all vertices have the same color, but in
contrast to the previous example, we do not have a free choice of color for each cycle. To
see this, consider R2 = (1, 3, 5)(2, 4, 6). The cycle (1, 3, 5) can be colored blue and (2, 4, 6)
yellow, or vice versa, but it is not possible to color all six vertices blue, or to color all
vertices yellow. Thus f (R2 ) = 2. If π = F = (2, 6)(3, 5)(1)(4), a ¬xed coloring is obtained
by choosing one of the cycles of length 2 and one of the cycles of length 1 to be colored
blue, thus producing 3 blue vertices. Consequently, f (F ) = 4. To obtain f (I), note that
all legal colorings are ¬xed by I, so f (I) is the number of colorings of 6 vertices with
exactly 3 blue and 3 yellow vertices, namely, ( 6 ) = 20. From the cycle decompositions
of the members of G, there are two permutations with f = 2, three with f = 4, and one
with f = 20; the others have f = 0. Thus the number of distinct colorings is

(2(2) + 3(4) + 20) = 3.

Problems For Section 5.3
1. Assume that two colorings of the vertices of a square are equivalent if one can be
mapped into the other by a permutation in the dihedral group G = D8 . If n colors
are available, ¬nd the number of distinct colorings.
2. In Problem 1, suppose that we color the sides of the square rather than the vertices.
Do we get the same answer?
3. In Problem 1, assume that only two colors are available, white and green. There
are 16 unrestricted colorings, but only 6 equivalence classes. List the equivalence
classes explicitly.
4. Consider a rigid rod lying on the x-axis from x = ’1 to x = 1, with three beads
attached. The beads are located at the endpoints (’1, 0) and (1, 0), and at the center
(0, 0). The beads are to be painted using n colors, and two colorings are regarded
as equivalent if one can be mapped into the other by a permutation in the group
G = {I, σ}, where σ is the 180 degree rotation about the vertical axis. Find the
number of distinct colorings.

5. In Problem 4, ¬nd the number of distinct colorings if the color of the central bead is
always black.
6. Consider the group of rotations of the regular tetrahedron (see Figure 5.3.2); G con-
sists of the following permutations.
(i) The identity;
(ii) Rotations by 120 degrees, clockwise or counterclockwise, about an axis through
a vertex and the opposite face. There are 8 such rotations (choose 1 of 4 vertices,
then choose a clockwise or counterclockwise direction);
(iii) Rotations by 180 degrees about the line joining the midpoints of two nontouching
edges. There are 3 such rotations.
Argue geometrically to show that there are no other rotations in the group, and show
that G is isomorphic to the alternating group A4 .

Ð bbb
ÐÐ bb
ÐÐ bb
4b • • • • •2

bb Ð
bb ÐÐÐ

Figure 5.3.2

7. Given n colors, ¬nd the number of distinct colorings of the vertices of a regular
tetrahedron, if colorings that can be rotated into each other are equivalent.
8. In Problem 7, assume that n = 4 and label the colors B,Y,W,G. Find the number of
distinct colorings if exactly two vertices must be colored B.
9. The group G of rotations of a cube consists of the following permutations of the
(i) The identity.
(ii) Rotations of ±90 or 180 degrees about a line through the center of two opposite
faces; there are 3 — 3 = 9 such rotations.
(iii) Rotations of ±120 degrees about a diagonal of the cube, i.e., a line joining two
opposite vertices (vertices that are a maximal distance apart). There are 4
diagonals, so there are 4 — 2 = 8 such rotations.
(iv) Rotations of 180 degrees about a line joining the midpoints of two opposite edges.
There are 6 such rotations. (An axis of rotation is determined by selecting one
of the four edges on the bottom of the cube, or one of the two vertical edges on
the front face.)
Argue geometrically to show that there are no other rotations in the group, and show
that G is isomorphic to the symmetric group S4 .

10. If six colors are available and each face of a cube is painted a di¬erent color, ¬nd the
number of distinct colorings.
11. Let G be the group of rotations of a regular p-gon, where p is an odd prime. If the
vertices of the p-gon are to be painted using at most n colors, ¬nd the number of
distinct colorings.
12. Use the result of Problem 11 to give an unusual proof of Fermat™s little theorem.

5.4 The Sylow Theorems
Considerable information about the structure of a ¬nite group G can be obtained by
factoring the order of G. Suppose that |G| = pr m where p is prime, r is a positive integer,
and p does not divide m. Then r is the highest power of p that divides the order of G.
We will prove, among other things, that G must have a subgroup of order pr , and any
two such subgroups must be conjugate. We will need the following result about binomial

5.4.1 Lemma
If n = pr m where p is prime, then ( pr ) ≡ m mod p. Thus if p does not divide m, then it
pr m
does not divide .

Proof. By the binomial expansion modulo p (see Section 3.4), which works for polynomials
as well as for ¬eld elements, we have
r r r r
(X + 1)p ≡ X p + 1p = X p + 1 mod p.

Raise both sides to the power m to obtain
(X + 1)n ≡ (X p + 1)m mod p.
r n m
On the left side, the coe¬cient of X p is ( pr ), and on the right side, it is ( m’1 ) = m.
The result follows. ™

5.4.2 De¬nitions and Comments
Let p be a prime number. The group G is said to be a p-group if the order of each element
of G is a power of p. (The particular power depends on the element.) If G is a ¬nite
group, then G is a p-group i¬ the order of G is a power of p. [The “if” part follows
from Lagrange™s theorem, and the “only if” part is a corollary to the Sylow theorems;
see (5.4.5).]
If |G| = pr m, where p does not divide m, then a subgroup P of G of order pr is called
a Sylow p-subgroup of G. Thus P is a p-subgroup of G of maximum possible size.

5.4.3 The Sylow Theorems
Let G be a ¬nite group of order pr m, where p is prime, r is a positive integer, and p does
not divide m. Then

(1) G has at least one Sylow p-subgroup, and every p-subgroup of G is contained in a
Sylow p-subgroup.
(2) Let np be the number of Sylow p-subgroups of G. Then np ≡ 1 mod p and np di-
vides m.
(3) All Sylow p-subgroups are conjugate. Thus if we de¬ne an equivalence relation on
subgroups by H ∼ K i¬ H = gKg ’1 for some g ∈ G, then the Sylow p-subgroups
comprise a single equivalence class. [Note that the conjugate of a Sylow p-subgroup
is also a Sylow p-subgroup, since it has the same number of elements pr .]

(1) Let G act on subsets of G of size pr by left multiplication. The number of
such subsets is ppr , which is not divisible by p by (5.4.1). Consequently, since orbits

partition the set acted on by the group, there is at least one subset S whose orbit size
is not divisible by p. If P is the stabilizer of S, then by (5.2.3), the size of the orbit
is [G : P ] = |G|/|P | = pr m/|P |. For this to fail to be divisible by p, we must have
pr ||P |, and therefore pr ¤ |P |. But for any ¬xed x ∈ S, the map of P into S given by
g ’ gx is injective. (The map is indeed into S because g belongs to the stabilizer of S,
so that gS = S.) Thus |P | ¤ |S| = pr . We conclude that |P | = pr , hence P is a Sylow
So far, we have shown that a Sylow p-subgroup P exists, but not that every p-subgroup
is contained in a Sylow p-subgroup. We will return to this in the course of proving (2)
and (3).
(2) and (3) Let X be the set of all Sylow p-subgroups of G. Then |X| = np and P
acts on X by conjugation, i.e., g • Q = gQg ’1 , g ∈ P . By (5.2.3), the size of any orbit
divides |P | = pr , hence is a power of p. Suppose that there is an orbit of size 1, that
is, a Sylow p-subgroup Q ∈ X such that gQg ’1 = Q, and therefore gQ = Qg, for every
g ∈ P . (There is at least one such subgroup, namely P .) Then P Q = QP , so by (1.3.6),
P Q = P, Q, , the subgroup generated by P and Q. Since |P | = |Q| = pr it follows from
(5.2.4) that |P Q| is a power of p, say pc . We must have c ¤ r because P Q is a subgroup
of G (hence |P Q| divides |G|). Thus

pr = |P | ¤ |P Q| ¤ pr , so |P | = |P Q| = pr .

But P is a subset of P Q, and since all sets are ¬nite, we conclude that P = P Q, and
therefore Q ⊆ P . Since both P and Q are of size pr , we have P = Q. Thus there is only
one orbit of size 1, namely {P }. Since by (5.2.3), all other orbit sizes are of the form pc
where c ≥ 1, it follows that np ≡ 1 mod p.
Now let R be a p-subgroup of G, and let R act by multiplication on Y , the set of
left cosets of P . Since |Y | = [G : P ] = |G|/|P | = pr m/pr = m, p does not divide |Y |.
Therefore some orbit size is not divisible by p. By (5.2.3), every orbit size divides |R|,
hence is a power of p. (See (5.4.5) below. We are not going around in circles because
(5.4.4) and (5.4.5) only depend on the existence of Sylow subgroups, which we have already

established.) Thus there must be an orbit of size 1, say {gP } with g ∈ G. If h ∈ R then
hgP = gP , that is, g ’1 hg ∈ P , or equally well, h ∈ gP g ’1 . Consequently, R is contained
in a conjugate of P . If R is a Sylow p-subgroup to begin with, then R is a conjugate of
P , completing the proof of (1) and (3).
To ¬nish (2), we must show that np divides m. Let G act on subgroups by conjugation.
The orbit of P has size np by (3), so by (5.2.3), np divides |G| = pr m. But p cannot be
a prime factor of np , since np ≡ 1 mod p. It follows that np must divide m. ™

5.4.4 Corollary (Cauchy™s Theorem)
If the prime p divides the order of G, then G has an element of order p.

Proof. Let P be a Sylow p-subgroup of G, and pick x ∈ P with x = 1. The order of x is
a power of p, say |x| = pk . Then xp has order p. ™

5.4.5 Corollary
The ¬nite group G is a p-group if and only if the order of G is a power of p.

Proof. If the order of G is not a power of p, then it is divisible by some other prime q. But
in this case, G has a Sylow q-subgroup, and therefore by (5.4.4), an element of order q.
Thus G cannot be a p-group. The converse was done in (5.4.2). ™

Problems For Section 5.4
1. Under the hypothesis of the Sylow theorems, show that G has a subgroup of index np .
2. Let P be a Sylow p-subgroup of the ¬nite group G, and let Q be any p-subgroup. If Q
is contained in the normalizer NG (P ), show that P Q is a p-subgroup.
3. Continuing Problem 2, show that Q is contained in P .
4. Let P be a Sylow p-subgroup of the ¬nite group G, and let H be a subgroup of G that
contains the normalizer NG (P ).
(a) If g ∈ NG (H), show that P and gP g ’1 are Sylow p-subgroups of H, hence they
are conjugate in H.
(b) Show that NG (H) = H.
5. Let P be a Sylow p-subgroup of the ¬nite group G, and let N be a normal subgroup
of G. Assume that p divides |N | and |G/N |, so that N and G/N have Sylow p-
subgroups. Show that [P N : P ] and p are relatively prime, and then show that P © N
is a Sylow p-subgroup of N .
6. Continuing Problem 5, show that P N/N is a Sylow p-subgroup of G/N .
7. Suppose that P is the unique Sylow p-subgroup of G. [Equivalently, P is a normal
Sylow p-subgroup of G; see (5.5.4).] Show that for each automorphism f of G, we have
f (P ) = P . [Thus P is a characteristic subgroup of G; see (5.7.1).]

8. The Sylow theorems are about subgroups whose order is a power of a prime p. Here is
a result about subgroups of index p. Let H be a subgroup of the ¬nite group G, and
assume that [G : H] = p. Let N be a normal subgroup of G such that N ¤ H and
[G : N ] divides p! (see Section 5.1, Problem 8). Show that [H : N ] divides (p ’ 1)!.
9. Continuing Problem 8, let H be a subgroup of the ¬nite group G, and assume that H
has index p, where p is the smallest prime divisor of |G|. Show that H G.

5.5 Applications Of The Sylow Theorems
The Sylow theorems are of considerable assistance in the problem of classifying, up to
isomorphism, all ¬nite groups of a given order n. But in this area, proofs tend to involve
intricate combinatorial arguments, best left to specialized texts in group theory. We will
try to illustrate some of the basic ideas while keeping the presentation clean and crisp.

5.5.1 De¬nitions and Comments
A group G is simple if G = {1} and the only normal subgroups of G are G itself and {1}.
We will see later that simple groups can be regarded as building blocks for arbitrary ¬nite
groups. Abelian simple groups are already very familiar to us; they are the cyclic groups
of prime order. For if x ∈ G, x = 1, then by simplicity (and the fact that all subgroups of
an abelian group are normal), G = x . If G is not of prime order, then G has a nontrivial
proper subgroup by (1.1.4), so G cannot be simple.
The following results will be useful.

5.5.2 Lemma
If H and K are normal subgroups of G and the intersection of H and K is trivial (i.e., {1}),
then hk = kh for every h ∈ H and k ∈ K.

Proof. We did this in connection with direct products; see the beginning of the proof
of (1.5.2). ™

5.5.3 Proposition
If P is a nontrivial ¬nite p-group, then P has a nontrivial center.

Proof. Let P act on itself by conjugation; see (5.1.3) and (5.2.2), Example 3. The orbits
are the conjugacy classes of P . The element x belongs to an orbit of size 1 i¬ x is in
the center Z(P ), since gxg ’1 = x for all g ∈ P i¬ gx = xg for all g ∈ P i¬ x ∈ Z(P ).
By the orbit-stabilizer theorem, an orbit size that is greater than 1 must divide |P |, and
therefore must be a positive power of p. If Z(P ) = {1}, then we have one orbit of size 1,
with all other orbit sizes ≡ 0 mod p. Thus |P | ≡ 1 mod p, contradicting the assumption
that P is a nontrivial p-group. ™

5.5.4 Lemma
P is a normal Sylow p-subgroup of G if and only if P is the unique Sylow p-subgroup
of G.

Proof. By Sylow (3), the Sylow p-subgroups form a single equivalence class of conjugate
subgroups. This equivalence class consists of a single element {P } i¬ gP g ’1 = P for
every g ∈ G , that is, i¬ P G. ™

5.5.5 Proposition
Let G be a ¬nite, nonabelian simple group. If the prime p divides the order of G, then
the number np of Sylow p-subgroups of G is greater than 1.

Proof. If p is the only prime divisor of |G|, then G is a nontrivial p-group, hence Z(G)
is nontrivial by (5.5.3). Since Z(G) G (see (5.1.3), Example 3), Z(G) = G, so that G
is abelian, a contradiction. Thus |G| is divisible by at least two distinct primes, so if P
is a Sylow p-subgroup, then {1} < P < G. If np = 1, then there is a unique Sylow
p-subgroup P , which is normal in G by (5.5.4). This contradicts the simplicity of G, so
we must have np > 1. ™

We can now derive some properties of groups whose order is the product of two distinct

5.5.6 Proposition
Let G be a group of order pq, where p and q are distinct primes.

(i) If q ≡ 1 mod p, then G has a normal Sylow p-subgroup.
(ii) G is not simple.
(iii) If p ≡ 1 mod q and q ≡ 1 mod p, then G is cyclic.

Proof. (i) By Sylow (2), np ≡ 1 mod p and np |q, so np = 1. The result follows from
(ii) We may assume without loss of generality that p > q. Then p cannot divide q ’ 1,
so q ≡ 1 mod p. By (i), G has a normal Sylow p-subgroup, so G is not simple.
(iii) By (i), G has a normal Sylow p-subgroup P and a normal Sylow q-subgroup Q.
Since P and Q are of prime order (p and q, respectively), they are cyclic. If x generates P
and y generates Q, then xy = yx by (5.5.2). [P and Q have trivial intersection because
any member of the intersection has order dividing both p and q.] But then xy has order
pq = |G| (see Section 1.1, Problem 8). Thus G = xy . ™

We now look at the more complicated case |G| = p2 q. The combinatorial argument in
the next proof is very interesting.

5.5.7 Proposition
Suppose that the order of the ¬nite group G is p2 q, where p and q are distinct primes.
Then G has either a normal Sylow p-subgroup or a normal Sylow q-subgroup. Thus G is
not simple.

Proof. If the conclusion is false then np and nq are both greater than 1. By Sylow (2), nq
divides p2 , so nq = p or p2 , and we will show that the second case leads to a contradiction.
A Sylow q-subgroup Q is of order q and is therefore cyclic. Furthermore, every element
of Q except the identity is a generator of Q. Conversely, any element of order q generates
a Sylow q-subgroup. Since the only divisors of q are 1 and q, any two distinct Sylow
q-subgroups have trivial intersection. Thus the number of elements of G of order q is
exactly nq (q ’ 1). If nq = p2 , then the number of elements that are not of order q is

p2 q ’ p2 (q ’ 1) = p2 .

Now let P be any Sylow p-subgroup of G. Then |P | = p2 , so no element of P can
have order q (the orders must be 1, p or p2 ). Since there are only p2 elements of order
unequal to q available, P takes care of all of them. Thus there cannot be another Sylow p-
subgroup, so np = 1, a contradiction. We conclude that nq must be p. Now by Sylow (2),
nq ≡ 1 mod q, hence p ≡ 1 mod q, so p > q. But np divides q, a prime, so np = q. Since
np ≡ 1 mod p, we have q ≡ 1 mod p, and consequently q > p. Our original assumption
that both np and nq are greater than one has led inexorably to a contradiction. ™

Problems For Section 5.5
1. Show that every group of order 15 is cyclic.
2. If G/Z(G) is cyclic, show that G = Z(G), and therefore G is abelian.
3. Show that for prime p, every group of order p2 is abelian.
4. Let G be a group with |G| = pqr, where p, q and r are distinct primes and (without
loss of generality) p > q > r. Show that |G| ≥ 1 + np (p ’ 1) + nq (q ’ 1) + nr (r ’ 1).
5. Continuing Problem 4, if G is simple, show that np , nq and nr are all greater than 1.
Then show that np = qr, nq ≥ p and nr ≥ q.
6. Show that a group whose order is the product of three distinct primes is not simple.
7. Let G be a simple group of order pr m, where r ≥ 1, m > 1, and the prime p does
not divide m. Let n = np be the number of Sylow p-subgroups of G. If H = NG (P ),
where P is a Sylow p-subgroup of G, then [G : H] = n (see Problem 1 of Section 5.4).
Show that P cannot be normal in G (hence n > 1), and conclude that |G| must
divide n!.
8. If G is a group of order 250, 000 = 24 56 , show that G is not simple.

5.6 Composition Series
5.6.1 De¬nitions and Comments
One way to break down a group into simpler components is via a subnormal series

1 = G0 G1 Gr = G.

“Subnormal” means that each subgroup Gi is normal in its successor Gi+1 . In a normal
series, the Gi are required to be normal subgroups of the entire group G. For convenience,
the trivial subgroup {1} will be written as 1.
Suppose that Gi is not a maximal normal subgroup of Gi+1 , equivalently (by the
correspondence theorem) Gi+1 /Gi is not simple. Then the original subnormal series can
be re¬ned by inserting a group H such that Gi H Gi+1 . We can continue re¬ning in the
hope that the process will terminate (it always will if G is ¬nite). If all factors Gi+1 /Gi
are simple, we say that the group G has a composition series. [By convention, the trivial
group has a composition series, namely {1} itself.]
The Jordan-H¨lder theorem asserts that if G has a composition series, the resulting
composition length r and the composition factors Gi+1 /Gi are unique (up to isomorphism
and rearrangement). Thus all re¬nements lead to essentially the same result. Simple
groups therefore give important information about arbitrary groups; if G1 and G2 have
di¬erent composition factors, they cannot be isomorphic.
Here is an example of a composition series. Let S4 be the group of all permutations
of {1, 2, 3, 4}, and A4 the subgroup of even permutations (normal in S4 by Section 1.3,
Problem 6). Let V be the four group (Section 1.2, Problem 6; normal in A4 , in fact in S4 ,
by direct veri¬cation). Let Z2 be any subgroup of V of order 2. Then

1 Z2 V A 4 S4 .

The proof of the Jordan-H¨lder theorem requires some technical machinery.

5.6.2 Lemma
H ¤ G and f is a homomorphism on G, then f (K)
(i) If K f (H).
H ¤ G and N
(ii) If K G, then N K N H.
D, then A(B © C)
(iii) If A, B, C and D are subgroups of G with A B and C
A(B © D), and by symmetry, C(D © A) C(D © B).
(iv) In (iii), A(B © C) © B © D = C(D © A) © D © B.
Equivalently, A(B © C) © D = C(D © A) © B.

(i) For h ∈ H, k ∈ K, we have f (h)f (k)f (h)’1 = f (hkh’1 ) ∈ f (K).
(ii) Let f be the canonical map of G onto G/N . By (i) we have N K/N N H/N .
The result follows from the correspondence theorem.
(iii) Apply (ii) with G = B, N = A, K = B © C, H = B © D.
(iv) The two versions are equivalent because A(B © C) ¤ B and C(D © A) ¤ D. If x
belongs to the set on the left, then x = ac for some a ∈ A, c ∈ B © C, and x also belongs

to D. But x = c(c’1 ac) = ca— for some a— ∈ A B. Since x ∈ D and c ∈ C ¤ D,
— — —
we have a ∈ D, hence a ∈ D © A. Thus x = ca ∈ C(D © A), and since x = ac, with
a ∈ A ¤ B and c ∈ B © C ¤ B, x ∈ C(D © A) © B. Therefore the left side is a subset of
the right side, and a symmetrical argument completes the proof. ™

The diagram below is helpful in visualizing the next result.


To keep track of symmetry, take mirror images about the dotted line. Thus the group A
will correspond to C, B to D, A(B © C) to C(D © A), and A(B © D) to C(D © B).

5.6.3 Zassenhaus Lemma
Let A, B, C and D be subgroups of G, with A B and C D. Then

A(B © D) ∼ C(D © B)
A(B © C) C(D © A)

Proof. By part (iii) of (5.6.2), the quotient groups are well-de¬ned. An element of the
group on the left is of the form ayA(B © C), a ∈ A, y ∈ B © D. But ay = y(y ’1 ay) = ya— ,
a— ∈ A. Thus ayA(B © C) = ya— A(B © C) = yA(B © C). Similarly, an element of the
right side is of the form zC(D © A) with z ∈ D © B = B © D. Thus if y, z ∈ B © D, then

yA(B © C) = zA(B © C) i¬ z ’1 y ∈ A(B © C) © B © D

and by part (iv) of (5.6.2), this is equivalent to

z ’1 y ∈ C(D © A) © D © B i¬ yC(D © A) = zC(D © A).

Thus if h maps yA(B © C) to yC(D © A), then h is a well-de¬ned bijection from the left
to the right side of Zassenhaus™ equation. By de¬nition of multiplication in a quotient
group, h is an isomorphism. ™

5.6.4 De¬nitions and Comments
If a subnormal series is re¬ned by inserting H between Gi and Gi+1 , let us allow H to
coincide with Gi or Gi+1 . If all such insertions are strictly between the “endgroups”, we
will speak of a proper re¬nement. Two series are equivalent if they have the same length
and their factor groups are the same, up to isomorphism and rearrangement.

5.6.5 Schreier Re¬nement Theorem
Let 1 = H0 H1 · · · Hr = G and 1 = K0 K1 · · · Ks = G be two subnormal
series for the group G. Then the series have equivalent re¬nements.

Proof. Let Hij = Hi (Hi+1 © Kj ), Kij = Kj (Kj+1 © Hi ). By Zassenhaus we have
Hi,j+1 ∼ Ki+1,j
Hij Kij
(In (5.6.3) take A = Hi , B = Hi+1 , C = Kj , D = Kj+1 ). We can now construct equivalent
re¬nements; the easiest way to see this is to look at a typical concrete example. The ¬rst
re¬nement will have r blocks of length s, and the second will have s blocks of length r.
Thus the length will be rs in both cases. With r = 2 and s = 3, we have

1 = H00 H01 H02 H03 = H1 = H10 H11 H12 H13 = H2 = G,
1 = K00 K10 K20 = K1 = K01 K11 K21 = K2 = K02 K12 K22 = K3 = G.

The corresponding factor groups are

H01 /H00 ∼ K10 /K00 , H02 /H01 ∼ K11 /K01 , H03 /H02 ∼ K12 /K02
= = =
H11 /H10 ∼ K20 /K10 , H12 /H11 ∼ K21 /K11 , H13 /H12 ∼ K22 /K12 .
= = =

(Notice the pattern; in each isomorphism, the ¬rst subscript in the numerator is increased
by 1 and the second subscript is decreased by 1 in going from left to right. The subscripts
in the denominator are unchanged.) The factor groups of the second series are a reordering
of the factor groups of the ¬rst series. ™
The hard work is now accomplished, and we have everything we need to prove the
main result.

5.6.6 Jordan-H¨lder Theorem
If G has a composition series S (in particular if G is ¬nite), then any subnormal se-
ries R without repetition can be re¬ned to a composition series. Furthermore, any two
composition series for G are equivalent.
Proof. By (5.6.5), R and S have equivalent re¬nements. Remove any repetitions from
the re¬nements to produce equivalent re¬nements R0 and S0 without repetitions. But a
composition series has no proper re¬nements, hence S0 = S, proving the ¬rst assertion.
If R is also a composition series, then R0 = R as well, and R is equivalent to S. ™

Problems For Section 5.6
1. Show that if G has a composition series, so does every normal subgroup of G.
2. Give an example of a group that has no composition series.
3. Give an example of two nonisomorphic groups with the same composition factors, up
to rearrangement.

Problems 4“9 will prove that the alternating group An is simple for all n ≥ 5. (A1
and A2 are trivial and hence not simple; A4 is not simple by the example given in (5.6.1);
A3 is cyclic of order 3 and is therefore simple.) In these problems, N stands for a normal
subgroup of An .

4. Show that if n ≥ 3, then An is generated by 3-cycles.
5. Show that if N contains a 3-cycle, then it contains all 3-cycles, so that N = An .
6. ¿From now on, assume that N is a proper normal subgroup of An , and n ≥ 5. Show
that no permutation in N contains a cycle of length 4 or more.
7. Show that no permutation in N contains the product of two disjoint 3-cycles. Thus
in view of Problems 4,5 and 6, every member of N is the product of an even number
of disjoint transpositions.
8. In Problem 7, show that the number of transpositions in a nontrivial member of N
must be at least 4.
9. Finally, show that the assumption that N contains a product of 4 or more disjoint
transpositions leads to a contradiction, proving that N = 1, so that An is simple. It
follows that a composition series for Sn is 1 An Sn .
10. A chief series is a normal series without repetition that cannot be properly re¬ned
to another normal series. Show that if G has a chief series, then any normal series
without repetition can be re¬ned to a chief series. Furthermore, any two chief series
of a given group are equivalent.
11. In a composition series, the factor groups Gi+1 /Gi are required to be simple. What
is the analogous condition for a chief series?

5.7 Solvable And Nilpotent Groups
Solvable groups are so named because of their connection with solvability of polynomial
equations, a subject to be explored in the next chapter. To get started, we need a property
of subgroups that is stronger than normality.

5.7.1 De¬nitions and Comments
A subgroup H of the group G is characteristic (in G) if for each automorphism f of G,
f (H) = H. Thus f restricted to H is an automorphism of H. Consequently, if H is
characteristic in G, then it is normal in G. If follows from the de¬nition that if H is
characteristic in K and K is characteristic in G, then H is characteristic in G. Another
useful result is the following.

(1) If H is characteristic in K and K is normal in G, then H is normal in G.

To see this, observe that any inner automorphism of G maps K to itself, so restricts
to an automorphism (not necessarily inner) of K. Further restriction to H results in an
automorphism of H, and the result follows.

5.7.2 More De¬nitions and Comments
The commutator subgroup G of a group G is the subgroup generated by all commu-
tators [x, y] = xyx’1 y ’1 . (Since [x, y]’1 = [y, x], G consists of all ¬nite products of
commutators.) Here are some basic properties.

(2) G is characteristic in G.
This follows because any automorphism f maps a commutator to a commutator:
f [x, y] = [f (x), f (y)].
(3) G is abelian if and only if G is trivial.
This holds because [x, y] = 1 i¬ xy = yx.
(4) G/G is abelian. Thus forming the quotient of G by G , sometimes called modding
out by G , in a sense “abelianizes” the group.
For G xG y = G yG x i¬ G xy = G yx i¬ xy(yx)’1 ∈ G i¬ xyx’1 y ’1 ∈ G , and this
holds for all x and y by de¬nition of G .
G, then G/N is abelian if and only if G ¤ N .
(5) If N
The proof of (4) with G replaced by N shows that G/N is abelian i¬ all commutators
belong to N , that is, i¬ G ¤ N .
The process of taking commutators can be iterated:

G(0) = G, G(1) = G , G(2) = (G ) ,

and in general,

G(i+1) = (G(i) ) , i = 0, 1, 2, . . . .

Since G(i+1) is characteristic in G(i) , an induction argument shows that each G(i) is
characteristic, hence normal, in G.
The group G is said to be solvable if G(r) = 1 for some r. We then have a normal

1 = G(r) G(r’1) G(0) = G

called the derived series of G.
Every abelian group is solvable, by (3). Note that a group that is both simple and
solvable must be cyclic of prime order. For the normal subgroup G must be trivial; if it
were G, then the derived series would never reach 1. By (3), G is abelian, and by (5.5.1),
G must be cyclic of prime order.
A nonabelian simple group G (such as An , n ≥ 5) cannot be solvable. For if G is
nonabelian, then G is not trivial. Thus G = G, and as in the previous paragraph, the
derived series will not reach 1.
There are several equivalent ways to describe solvability.

5.7.3 Proposition
The following conditions are equivalent.
(i) G is solvable.
(ii) G has a normal series with abelian factors.

(iii) G has a subnormal series with abelian factors.

Proof. Since (i) implies (ii) by (4) and (ii) implies (iii) by de¬nition of normal and sub-
normal series, the only problem is (iii) implies (i). Suppose G has a subnormal series

1 = Gr Gr’1 G1 G0 = G

with abelian factors. Since G/G1 is abelian, we have G ¤ G1 by (5), and an induction
argument then shows that G(i) ¤ Gi for all i. [The inductive step is G(i+1) = (G(i) ) ¤
Gi ¤ Gi+1 since Gi /Gi+1 is abelian.] Thus G(r) ¤ Gr = 1. ™

The next result gives some very useful properties of solvable groups.

5.7.4 Proposition
Subgroups and quotients of a solvable group are solvable. Conversely, if N is normal
subgroup of G and both N and G/N are solvable, then G is solvable.

Proof. If H is a subgroup of the solvable group G, then H is solvable because H (i) ¤ G(i)
for all i. If N is a normal subgroup of the solvable group G, observe that commutators of
G/N look like xyx’1 y ’1 N , so (G/N ) = G N/N . (Not G /N , since N is not necessarily
a subgroup of G .) Inductively,

(G/N )(i) = G(i) N/N

and since N/N is trivial, G/N is solvable. Conversely, suppose that we have a subnormal
series from N0 = 1 to Nr = N , and a subnormal series from G0 /N = 1 (i.e., G0 = N )
to Gs /N = G/N (i.e., Gs = G) with abelian factors in both cases. Then we splice the
series of Ni ™s to the series of Gi ™s. The latter series is subnormal by the correspondence
theorem, and the factors remain abelian by the third isomorphism theorem. ™

5.7.5 Corollary
If G has a composition series, in particular if G is ¬nite, then G is solvable if and only if
the composition factors of G are cyclic of prime order.

Proof. Let Gi+1 /Gi be a composition factor of the solvable group G. By (5.7.4), Gi+1 is
solvable, and again by (5.7.4), Gi+1 /Gi is solvable. But a composition factor must be a
simple group, so Gi+1 /Gi is cyclic of prime order, as observed in (5.7.2). Conversely, if
the composition factors of G are cyclic of prime order, then the composition series is a
subnormal series with abelian factors. ™

Nilpotent groups arise from a di¬erent type of normal series. We will get at this idea
indirectly, and give an abbreviated treatment.

5.7.6 Proposition
If G is a ¬nite group, the following conditions are equivalent, and de¬ne a nilpotent group.
[Nilpotence of an arbitrary group will be de¬ned in (5.7.8).]

(a) G is the direct product of its Sylow subgroups.
(b) Every Sylow subgroup of G is normal.

Proof. (a) implies (b): By (1.5.3), the factors of a direct product are normal subgroups.
(b) implies (a): By (5.5.4), there is a unique Sylow pi -subgroup Hi for each prime
divisor pi of |G|, i = 1, . . . , k. By successive application of (5.2.4), we have |H1 · · · Hk | =
|H1 | · · · |Hk |, which is |G| by de¬nition of Sylow p-subgroup. Since all sets are ¬nite,
G = H1 · · · Hk . Furthermore, each Hi © j=i Hj is trivial, because the orders of the Hi
are powers of distinct primes. By (1.5.4), G is the direct product of the Hi . ™

5.7.7 Corollary
Every ¬nite abelian group and every ¬nite p-group is nilpotent.

Proof. A ¬nite abelian group must satisfy condition (b) of (5.7.6). If P is a ¬nite p-
group, then P has only one Sylow subgroup, P itself, so the conditions of (5.7.6) are
automatically satis¬ed. ™

We now connect this discussion with normal series. Suppose that we are trying to
build a normal series for the group G, starting with G0 = 1. We take G1 to be Z(G), the
center of G; we have G1 G by (5.1.3), Example 3. We de¬ne G2 by the correspondence

G2 /G1 = Z(G/G1 )

and since Z(G/G1 ) G/G1 , we have G2 G. In general, we take

Gi /Gi’1 = Z(G/Gi’1 ),

and by induction we have Gi G. The di¬culty is that there is no guarantee that Gi will
ever reach G. However, we will succeed if G is a ¬nite p-group. The key point is that a
nontrivial ¬nite p-group has a nontrivial center, by (5.5.3). Thus by induction, Gi /Gi’1
is nontrivial for every i, so Gi’1 < Gi . Since G is ¬nite, it must eventually be reached.

5.7.8 De¬nitions and Comments
A central series for G is a normal series 1 = G0 G1 Gr = G such that
Gi /Gi’1 ⊆ Z(G/Gi’1 ) for every i = 1, . . . , r. (The series just discussed is a special
case called the upper central series.) An arbitrary group G is said to be nilpotent if it
has a central series. Thus a ¬nite p-group is nilpotent, and in particular, every Sylow
p-subgroup is nilpotent. Now a direct product of a ¬nite number of nilpotent groups is
nilpotent. (If Gij is the ith term of a central series of the j th factor Hj , with Gij = G
if the series has already terminated at G, then j Gij will be the ith term of a central

series for j Hj .) Thus a ¬nite group that satis¬es the conditions of (5.7.6) has a central
series. Conversely, it can be shown that a ¬nite group that has a central series satis¬es
(5.7.6), so the two de¬nitions of nilpotence agree for ¬nite groups.
Note that a nilpotent group is solvable. For if Gi /Gi’1 ⊆ Z(G/Gi’1 ), then the
elements of Gi /Gi’1 commute with each other since they commute with everything in
G/Gi’1 ; thus Gi /Gi’1 is abelian. Consequently, a ¬nite p-group is solvable.

Problems For Section 5.7
1. Give an example of a nonabelian solvable group.
2. Show that a solvable group that has a composition series must be ¬nite.
3. Prove directly (without making use of nilpotence) that a ¬nite p-group is solvable.
4. Give an example of a solvable group that is not nilpotent.
5. Show that if n ≥ 5, then Sn is not solvable.
6. If P is a ¬nite simple p-group, show that P has order p.
7. Let P be a nontrivial ¬nite p-group. Show that P has a normal subgroup N whose
index [P : N ] is p.
8. Let G be a ¬nite group of order pr m, where r is a positive integer and p does not
divide m. Show that for any k = 1, 2, . . . , r, G has a subgroup of order pk .
9. Give an example of a group G with a normal subgroup N such that N and G/N are
abelian, but G is not abelian. (If “abelian” is replaced by “solvable”, no such example
is possible, by (5.7.4).)
10. If G is a solvable group, its derived length, dl(G), is the smallest nonnegative integer r
such that G(r) = 1. If N is a normal subgroup of the solvable group G, what can be
said about the relation between dl(G), dl(N ) and dl(G/N )?

5.8 Generators And Relations
In (1.2.4) we gave an informal description of the dihedral group via generators and rela-
tions, and now we try to make the ideas more precise.

5.8.1 De¬nitions and Comments
The free group G on the set S (or the free group with basis S) consists of all words on S,
that is, all ¬nite sequences x1 · · · xn , n = 0, 1, . . . , where each xi is either an element
of S or the inverse of an element of S. We regard the case n = 0 as the empty word ».
The group operation is concatenation, subject to the constraint that if s and s’1 occur
in succession, they can be cancelled. The empty word is the identity, and inverses are
calculated in the only reasonable way, for example, (stu)’1 = u’1 t’1 s’1 . We say that G
is free on S.
Now suppose that G is free on S, and we attempt to construct a homomorphism f
from G to an arbitrary group H. The key point is that f is completely determined by its

values on S. If f (s1 ) = a, f (s2 ) = b, f (s3 ) = c, then

f (s1 s’1 s3 ) = f (s1 )f (s2 )’1 f (s3 ) = ab’1 c.

Here is the formal statement, followed by an informal proof.

5.8.2 Theorem
If G is free on S and g is an arbitrary function from S to a group H, then there is a
unique homomorphism f : G ’ H such that f = g on S.

Proof. The above discussion is a nice illustration of a concrete example with all the
features of the general case. The analysis shows both existence and uniqueness of f . A
formal proof must show that all aspects of the general case are covered. For example,
if u = s1 s’1 s3 and v = s1 s’1 s’1 s4 s3 , then f (u) = f (v), so that cancellation of s’1 s4
2 2 4 4
causes no di¬culty. Speci¬c calculations of this type are rather convincing, and we will
not pursue the formal details. (See, for example, Rotman, An Introduction to the Theory
of Groups, pp. 343“345.) ™

5.8.3 Corollary
Any group H is a homomorphic image of a free group.

Proof. Let S be a set of generators for H (if necessary, take S = H), and let G be free
on S. De¬ne g(s) = s for all s ∈ S. If f is the unique extension of g to G, then since S
generates H, f is an epimorphism. ™

Returning to (1.2.4), we described a group H using generators R and F , and relations
R = I, F 2 = I, RF = F R’1 . The last relation is equivalent to RF RF = I, since F 2 = I.

The words Rn , F 2 and RF RF are called relators, and the speci¬cation of generators and
relations is called a presentation. We use the notation

H = R, F | Rn , F 2 , RF RF

or the long form

H = R, F | Rn = I, F 2 = I, RF = F R’1 .

We must say precisely what it means to de¬ne a group by generators and relations, and
show that the above presentation yields a group isomorphic to the dihedral group D2n .
We start with the free group on {R, F } and set all relators equal to the identity. It is
natural to mod out by the subgroup generated by the relators, but there is a technical
di¬culty; this subgroup is not necessarily normal.

5.8.4 De¬nition
Let G be free on the set S, and let K be a subset of G. We de¬ne the group S | K as
G/K, where K is the smallest normal subgroup of G containing K.
Unfortunately, it is a theorem of mathematical logic that there is no algorithm which
when given a presentation, will ¬nd the order of the group. In fact, there is no algorithm
to determine whether a given word of S | K coincides with the identity. Logicians say
that the word problem for groups is unsolvable. But although there is no general solution,
there are speci¬c cases that can be analyzed, and the following result is very helpful.

5.8.5 Von Dyck™s Theorem
Let H = S | K be a presentation, and let L be a group that is generated by the words
in S. If L satis¬es all the relations of K, then there is an epimorphism ± : H ’ L.
Consequently, |H| ≥ |L|.

Proof. Let G be free on S, and let i be the identity map from S, regarded as a subset of G,
to S, regarded as a subset of L. By (5.8.2), i has a unique extension to a homomorphism
f of G into L, and in fact f is an epimorphism because S generates L. Now f maps
any word of G to the same word in L, and since L satis¬es all the relations, we have
K ⊆ ker f . But the kernel of f is a normal subgroup of G, hence K ⊆ ker f . The factor
theorem provides an epimorphism ± : G/K ’ L. ™

5.8.6 Justifying a presentation
If L is a ¬nite group generated by the words of S, then in practice, the crucial step in
identifying L with H = S | K is a proof that |H| ¤ |L|. If we can accomplish this,
then by (5.8.5), |H| = |L|. In this case, ± is a surjective map of ¬nite sets of the same
size, so ± is injective as well, hence is an isomorphism. For the dihedral group we have
H = F, R | Rn , F 2 , RF RF and L = D2n . In (1.2.4) we showed that each word of H can
be expressed as Ri F j with 0 ¤ i ¤ n’1 and 0 ¤ j ¤ 1. Therefore |H| ¤ 2n = |D2n | = |L|.
Thus the presentation H is a legitimate description of the dihedral group.

Problems For Section 5.8
1. Show that a presentation of the cyclic group of order n is a | an .
2. Show that the quaternion group (see (2.1.3, Example 4)) has a presentation a, b |
a4 = 1, b2 = a2 , ab = ba’1 .
3. Show that H = a, b | a3 = 1, b2 = 1, ba = a’1 b is a presentation of S3 .
4. Is the presentation of a group unique?

In Problems 5“11, we examine a di¬erent way of assembling a group from subgroups,
which generalizes the notion of a direct product. Let N be a normal subgroup of G,
and H an arbitrary subgroup. We say that G is the semidirect product of N by H if
G = N H and N © H = 1. (If H G, we have the direct product.) For notational
convenience, the letter n, possibly with subscripts, will always indicate a member of N ,

and similarly h will always belong to H. In Problems 5 and 6, we assume that G is the
semidirect product of N by H.
5. If n1 h1 = n2 h2 , show that n1 = n2 and h1 = h2 .
6. If i : N ’ G is inclusion and π : G ’ H is projection (π(nh) = h), then the sequence

i π
’ ’ ’H ’
1 N G 1

is exact. Note that π is well-de¬ned by Problem 5, and verify that π is a homomor-
phism. Show that the sequence splits on the right, i.e., there is a homomorphism
ψ : H ’ G such that π —¦ ψ = 1.
7. Conversely, suppose that the above exact sequence splits on the right. Since ψ is
injective, we can regard H (and N as well) as subgroups of G, with ψ and i as
inclusion maps. Show that G is the semidirect product of N by H.
8. Let N and H be arbitrary groups, and let f be a homomorphism of H into Aut N ,
the group of automorphisms of N . De¬ne a multiplication on G = N — H by

(n1 , h1 )(n2 , h2 ) = (n1 f (h1 )(n2 ), h1 h2 ).

[f (h1 )(n2 ) is the value of the automorphism f (h1 ) at the element n2 .] A lengthy but
straightforward calculation shows that G is a group with identity (1, 1) and inverses
given by (n, h)’1 = (f (h’1 )(n’1 ), h’1 ). Show that G is the semidirect product of
N — {1} by {1} — H.
9. Show that every semidirect product arises from the construction of Problem 8.
10. Show by example that it is possible for a short exact sequence of groups to split on
the right but not on the left.
[If h : G ’ N is a left-splitting map in the exact sequence of Problem 6, then h and π
can be used to identify G with the direct product of N and H. Thus a left-splitting
implies a right-splitting, but, unlike the result for modules in (4.7.4), not conversely.]
11. Give an example of a short exact sequence of groups that does not split on the right.
12. (The Frattini argument, frequently useful in a further study of group theory.) Let
N be a normal subgroup of the ¬nite group G, and let P be a Sylow p-subgroup of
N . If NG (P ) is the normalizer of P in G, show that G = NG (P )N (= N NG (P ) by
(1.4.3)).[If g ∈ G, look at the relation between P and gP g ’1 .]
13. Let N = {1, a, a2 , . . . , an’1 } be a cyclic group of order n, and let H = {1, b} be a
cyclic group of order 2. De¬ne f : H ’ Aut N by taking f (b) to be the automorphism
that sends a to a’1 . Show that the dihedral group D2n is the semidirect product of N
by H. (See Problems 8 and 9 for the construction of the semidirect product.)
14. In Problem 13, replace N by an in¬nite cyclic group

{. . . , a’2 , a’1 , 1, a, a2 , . . . }.

Give a presentation of the semidirect product of N by H. This group is called the
in¬nite dihedral group D∞ .

Concluding Remarks
Suppose that the ¬nite group G has a composition series

1 = G0 G1 · · · Gr = G.

If Hi = Gi /Gi’1 , then we say that Gi is an extension of Gi’1 by Hi in the sense that
Gi and Gi /Gi’1 ∼ Hi . If we were able to solve the extension problem (¬nd all
Gi’1 =
possible extensions of Gi’1 by Hi ) and we had a catalog of all ¬nite simple groups, then
we could build a catalog of all ¬nite groups. This sharpens the statement made in (5.6.1)
about the importance of simple groups.
Chapter 6

Galois Theory

6.1 Fixed Fields and Galois Groups
Galois theory is based on a remarkable correspondence between subgroups of the Galois
group of an extension E/F and intermediate ¬elds between E and F . In this section
we will set up the machinery for the fundamental theorem. [A remark on notation:
Throughout the chapter, the composition „ —¦ σ of two automorphisms will be written as
a product „ σ.]

6.1.1 De¬nitions and Comments
Let G = Gal(E/F ) be the Galois group of the extension E/F . If H is a subgroup of G,
the ¬xed ¬eld of H is the set of elements ¬xed by every automorphism in H, that is,

F(H) = {x ∈ E : σ(x) = x for every σ ∈ H}.

If K is an intermediate ¬eld, that is, F ¤ K ¤ E, de¬ne

G(K) = Gal(E/K) = {σ ∈ G : σ(x) = x for every x ∈ K}.

I like the term “¬xing group of K” for G(K), since G(K) is the group of automorphisms
of E that leave K ¬xed. Galois theory is about the relation between ¬xed ¬elds and ¬xing
groups. In particular, the next result suggests that the smallest sub¬eld F corresponds
to the largest subgroup G.

6.1.2 Proposition
Let E/F be a ¬nite Galois extension with Galois group G = Gal(E/F ). Then

(i) The ¬xed ¬eld of G is F ;
(ii) If H is a proper subgroup of G, then the ¬xed ¬eld of H properly contains F .


Proof. (i) Let F0 be the ¬xed ¬eld of G. If σ is an F -automorphism of E, then by
de¬nition of F0 , σ ¬xes everything in F0 . Thus the F -automorphisms of G coincide with
the F0 -automorphisms of G. Now by (3.4.7) and (3.5.8), E/F0 is Galois. By (3.5.9), the
size of the Galois group of a ¬nite Galois extension is the degree of the extension. Thus
[E : F ] = [E : F0 ], so by (3.1.9), F = F0 .
(ii) Suppose that F = F(H). By the theorem of the primitive element (3.5.12), we
have E = F (±) for some ± ∈ E. De¬ne a polynomial f (X) ∈ E[X] by

(X ’ σ(±)).
f (X) =

If „ is any automorphism in H, then we may apply „ to f (that is, to the coe¬cients of f ;
we discussed this idea in the proof of (3.5.2)). The result is

(X ’ („ σ)(±)).
(„ f )(X) =

But as σ ranges over all of H, so does „ σ, and consequently „ f = f . Thus each coe¬cient
of f is ¬xed by H, so f ∈ F [X]. Now ± is a root of f , since X ’ σ(±) is 0 when X = ±
and σ is the identity. We can say two things about the degree of f :
(1) By de¬nition of f , deg f = |H| < |G| = [E : F ], and, since f is a multiple of the
minimal polynomial of ± over F ,
(2) deg f ≥ [F (±) : F ] = [E : F ], and we have a contradiction. ™
There is a converse to the ¬rst part of (6.1.2).

6.1.3 Proposition
Let E/F be a ¬nite extension with Galois group G. If the ¬xed ¬eld of G is F , then E/F
is Galois.
Proof. Let G = {σ1 , . . . , σn }, where σ1 is the identity. To show that E/F is normal,
we consider an irreducible polynomial f ∈ F [X] with a root ± ∈ E. Apply each au-
tomorphism in G to ±, and suppose that there are r distinct images ± = ±1 = σ1 (±),
±2 = σ2 (±), . . . , ±r = σr (±). If σ is any member of G, then σ will map each ±i to some
±j , and since σ is an injective map of the ¬nite set {±1 , . . . , ±r } to itself, it is surjective as
well. To put it simply, σ permutes the ±i . Now we examine what σ does to the elementary
symmetric functions of the ±i , which are given by
e1 = ±i , e2 = ±i ±j , e3 = ±i ±j ±k , . . . ,
i=1 i<j i<j<k
er = ±i .

Since σ permutes the ±i , it follows that σ(ei ) = ei for all i. Thus the ei belong to the
¬xed ¬eld of G, which is F by hypothesis. Now we form a monic polynomial whose roots
are the ±i :
g(X) = (X ’ ±1 ) · · · (X ’ ±r ) = X r ’ e1 X r’1 + e2 X r’2 ’ · · · + (’1)r er .

Since the ei belong to F , g ∈ F [X], and since the ±i are in E, g splits over E. We claim
that g is the minimal polynomial of ± over F . To see this, let h(X) = b0 +b1 X +· · ·+bm X m
be any polynomial in F [X] having ± as a root. Applying σi to the equation

b0 + b 1 ± + · · · b m ± m = 0

we have

b0 + b1 ±i + · · · bm ±i = 0,

so that each ±i is a root of h, hence g divides h and therefore g =min(±, F ). But our
original polynomial f ∈ F [X] is irreducible and has ± as a root, so it must be a constant
multiple of g. Consequently, f splits over E, proving that E/F is normal. Since the ±i ,
i = 1, . . . r, are distinct, g has no repeated roots. Thus ± is separable over F , which shows
that the extension E/F is separable. ™

It is pro¬table to examine elementary symmetric functions in more detail.

6.1.4 Theorem
Let f be a symmetric polynomial in the n variables X1 , . . . , Xn . [This means that if σ is
any permutation in Sn and we replace Xi by Xσ(i) for i = 1, . . . , n, then f is unchanged.]
If e1 , . . . , en are the elementary symmetric functions of the Xi , then f can be expressed
as a polynomial in the ei .

Proof. We give an algorithm. The polynomial f is a linear combination of monomials
of the form X1 1 · · · Xnn , and we order the monomials lexicographically: X1 1 · · · Xnn >
r r
r r

X1 · · · Xnn i¬ the ¬rst disagreement between ri and si results in ri > si . Since f is
s1 s

symmetric, all terms generated by applying a permutation σ ∈ Sn to the subscripts of
X1 1 · · · Xnn will also contribute to f . The idea is to cancel the leading terms (those
r r

associated with the monomial that is ¬rst in the ordering) by subtracting an expression
of the form

et1 et2 · · · etn = (X1 + · · · + Xn )t1 · · · (X1 · · · Xn )tn

which has leading term

X11 (X1 X2 )t2 (X1 X2 X3 )t3 · · · (X1 · · · Xn )tn = X11 +···+tn X22 +···+tn · · · Xnn .
t t t t

This will be possible if we choose

t1 = r1 ’ r2 , t2 = r2 ’ r3 , . . . , tn’1 = rn’1 ’ rn , tn = rn .

After subtraction, the resulting polynomial has a leading term that is below X1 1 · · · Xnn
r r

in the lexicographical ordering. We can then repeat the procedure, which must terminate
in a ¬nite number of steps. ™

6.1.5 Corollary
If g is a polynomial in F [X] and f (±1 , . . . , ±n ) is any symmetric polynomial in the roots
±1 , . . . , ±n of g, then f ∈ F [X].
Proof. We may assume without loss of generality that g is monic. Then in a splitting
¬eld of g we have
g(X) = (X ’ ±1 ) · · · (X ’ ±n ) = X n ’ e1 X n’1 + · · · + (’1)n en .
By (6.1.4), f is a polynomial in the ei , and since the ei are simply ± the coe¬cients of g,
the coe¬cients of f are in F . ™

6.1.6 Dedekind™s Lemma
The result that the size of the Galois group of a ¬nite Galois extension is the degree of
the extension can be proved via Dedekind™s lemma, which is of interest in its own right.
Let G be a group and E a ¬eld. A character from G to E is a homomorphism from G
to the multiplicative group E — of nonzero elements of E. In particular, an automorphism
of E de¬nes a character with G = E — , as does a monomorphism of E into a ¬eld L.
Dedekind™s lemma states that if σ1 , . . . , σn are distinct characters from G to E, then the
σi are linearly independent over E. The proof is given in Problems 3 and 4.

Problems For Section 6.1
2 2 2
1. Express X1 X2 X3 + X1 X2 X3 + X1 X2 X3 in terms of elementary symmetric functions.
2 2 2 2 2 2
2. Repeat Problem 1 forX1 X2 + X1 X3 + X1 X2 + X1 X3 + X2 X3 + X2 X3 + 4X1 X2 X3 .
3. To begin the proof of Dedekind™s lemma, suppose that the σi are linearly dependent.
By renumbering the σi if necessary, we have
a1 σ1 + · · · ar σr = 0
where all ai are nonzero and r is as small as possible. Show that for every h and g ∈ G,
we have
ai σ1 (h)σi (g) = 0 (1)

ai σi (h)σi (g) = 0. (2)

[Equations (1) and (2) are not the same; in (1) we have σ1 (h), not σi (h).]
4. Continuing Problem 3, subtract (2) from (1) to get
ai (σ1 (h) ’ σi (h))σi (g) = 0. (3)

With g arbitrary, reach a contradiction by an appropriate choice of h.

5. If G is the Galois group of Q( 3 2) over Q, what is the ¬xed ¬eld of G?
6. Find the Galois group of C/R.
7. Find the ¬xed ¬eld of the Galois group of Problem 6.

6.2 The Fundamental Theorem
With the preliminaries now taken care of, we can proceed directly to the main result.

6.2.1 Fundamental Theorem of Galois Theory
Let E/F be a ¬nite Galois extension with Galois group G. If H is a subgroup of G,
let F(H) be the ¬xed ¬eld of H, and if K is an intermediate ¬eld, let G(K) be Gal(E/K),
the ¬xing group of K (see (6.1.1)).

(1) F is a bijective map from subgroups to intermediate ¬elds, with inverse G. Both maps
are inclusion-reversing, that is, if H1 ¤ H2 then F(H1 ) ≥ F(H2 ), and if K1 ¤ K2 ,
then G(K1 ) ≥ G(K2 ).
(2) Suppose that the intermediate ¬eld K corresponds to the subgroup H under the
Galois correspondence. Then
(a) E/K is always normal (hence Galois);
(b) K/F is normal if and only if H is a normal subgroup of G, and in this case,
(c) the Galois group of K/F is isomorphic to the quotient group G/H. Moreover,
whether or not K/F is normal,
(d) [K : F ] = [G : H] and [E : K] = |H|.
(3) If the intermediate ¬eld K corresponds to the subgroup H and σ is any automorphism
in G, then the ¬eld σK = {σ(x) : x ∈ K} corresponds to the conjugate subgroup
σHσ ’1 . For this reason, σK is called a conjugate sub¬eld of K.

The following diagram may aid the understanding.

| |
| |
F 1

As we travel up the left side from smaller to larger ¬elds, we move down the right side
from larger to smaller groups. A statement about K/F , an extension at the bottom of
the left side, corresponds to a statement about G/H, located at the top of the right side.
Similarly, a statement about E/K corresponds to a statement about H/1 = H.

Proof. (1) First, consider the composite mapping H ’ F(H) ’ GF(H). If σ ∈ H then σ
¬xes F(H) by de¬nition of ¬xed ¬eld, and therefore σ ∈ GF(H) = Gal(E/F(H)). Thus
H ⊆ GF(H). If the inclusion is proper, then by (6.1.2) part (ii) with F replaced by F(H),

we have F(H) > F(H), a contradiction. [Note that E/K is a Galois extension for any
intermediate ¬eld K, by (3.4.7) and (3.5.8).] Thus GF(H) = H.
Now consider the mapping K ’ G(K) ’ FG(K) = F Gal(E/K). By (6.1.2) part (i)
with F replaced by K, we have FG(K) = K. Since both F and G are inclusion-reversing
by de¬nition, the proof of (1) is complete.
(3) The ¬xed ¬eld of σHσ ’1 is the set of all x ∈ E such that σ„ σ ’1 (x) = x for every
„ ∈ H. Thus

F(σHσ ’1 ) = {x ∈ E : σ ’1 (x) ∈ F(H)} = σ(F(H)).

(2a) This was observed in the proof of (1).
(2b) If σ is an F -monomorphism of K into E, then by (3.5.2) and (3.5.6), σ extends
to an F -monomorphism of E into itself, in other words (see (3.5.6)), an F -automorphism
of E. Thus each such σ is the restriction to K of a member of G. Conversely, the
restriction of an automorphism in G to K is an F -monomorphism of K into E. By (3.5.5)
and (3.5.6), K/F is normal i¬ for every σ ∈ G we have σ(K) = K. But by (3), σ(K)
corresponds to σHσ ’1 and K to H. Thus K/F is normal i¬ σHσ ’1 = H for every σ ∈ G,
i.e., H G.
(2c) Consider the homomorphism of G = Gal(E/F ) to Gal(K/F ) given by σ ’ σ|K .
The map is surjective by the argument just given in the proof of (2b). The kernel is the
set of all automorphisms in G that restrict to the identity on K, that is, Gal(E/K) = H.
The result follows from the ¬rst isomorphism theorem.
(2d) By (3.1.9), [E : F ] = [E : K][K : F ]. The term on the left is |G| by (3.5.9), and
the ¬rst term on the right is | Gal(E/K)| by (2a), and this in turn is |H| since H = G(K).
Thus |G| = |H|[K : F ], and the result follows from Lagrange™s theorem. [If K/F is
normal, the proof is slightly faster. The ¬rst statement follows from (2c). To prove the
second, note that by (3.1.9) and (3.5.9),

[E : F ]
= |H|.] ™
[E : K] = =
[K : F ]

The next result is reminiscent of the second isomorphism theorem, and is best visu-
alized via the diamond diagram of Figure 6.2.1. In the diagram, EK is the composite of
the two ¬elds E and K, that is, the smallest ¬eld containing both E and K.

6.2.2 Theorem
Let E/F be a ¬nite Galois extension and K/F an arbitrary extension. Assume that E
and K are both contained in a common ¬eld, so that it is sensible to consider the com-
posite EK. Then

(1) EK/K is a ¬nite Galois extension;
(2) Gal(EK/K) is embedded in Gal(E/F ), where the embedding is accomplished by
restricting automorphisms in Gal(EK/K) to E;
(3) The embedding is an isomorphism if and only if E © K = F .

EK q
xx qq
xx qq
xx qq
E qq K
qq ww
qq ww
qq ww
q ww


Figure 6.2.1

Proof. (1) By the theorem of the primitive element (3.5.12), we have E = F [±] for some
± ∈ E, so EK = KF [±] = K[±]. The extension K[±]/K is ¬nite because ± is algebraic
over F , hence over K. Since ±, regarded as an element of EK, is separable over F and
hence over K, it follows that EK/K is separable. [To avoid breaking the main line of
thought, this result will be developed in the exercises (see Problems 1 and 2).]
Now let f be the minimal polynomial of ± over F , and g the minimal polynomial of ±
over K. Since f ∈ K[X] and f (±) = 0, we have g | f , and the roots of g must belong to
E ⊆ EK = K[±] because E/F is normal. Therefore K[±] is a splitting ¬eld for g over K,
so by (3.5.7), K[±]/K is normal.
(2) If σ is an automorphism in Gal(EK/K), restrict σ to E, thus de¬ning a homomor-
phism from Gal(EK/K) to Gal(E/F ). (Note that σ|E is an automorphism of E because
E/F is normal.) Now σ ¬xes K, and if σ belongs to the kernel of the homomorphism,
then σ also ¬xes E, so σ ¬xes EK = K[±]. Thus σ is the identity, and the kernel is trivial,
proving that the homomorphism is actually an embedding.
(3) The embedding of (2) maps Gal(EK/K) to a subgroup H of Gal(E/F ), and we
will ¬nd the ¬xed ¬eld of H. By (6.1.2), the ¬xed ¬eld of Gal(EK/K) is K, and since
the embedding just restricts automorphisms to E, the ¬xed ¬eld of H must be E © K.
By the fundamental theorem, H = Gal(E/(E © K)). Thus

H = Gal(E/F ) i¬ Gal(E/(E © K)) = Gal(E/F ),

and by applying the ¬xed ¬eld operator F, we see that this happens if and only if E ©
K = F. ™

Problems For Section 6.2
1. Let E = F (±1 , . . . , ±n ), where each ±i is algebraic and separable over F . We are going
to show that E is separable over F . Without loss of generality, we can assume that the
characteristic of F is a prime p, and since F/F is separable, the result holds for n = 0.
To carry out the inductive step, let Ei = F (±1 , . . . , ±i ), so that Ei+1 = Ei (±i+1 ).
Show that Ei+1 = Ei (Ei+1 ). (See Section 3.4, Problems 4“8, for the notation.)
2. Continuing Problem 1, show that E is separable over F .

3. Let E = F (±1 , . . . , ±n ), where each ±i is algebraic over F . If for each i = 1, . . . , n, all
the conjugates of ±i (the roots of the minimal polynomial of ±i over F ) belong to E,
show that E/F is normal.
4. Suppose that F = K0 ¤ K1 ¤ · · · ¤ Kn = E, where E/F is a ¬nite Galois extension,
and that the intermediate ¬eld Ki corresponds to the subgroup Hi under the Galois
correspondence. Show that Ki /Ki’1 is normal (hence Galois) if and only if Hi Hi’1 ,
and in this case, Gal(Ki /Ki’1 ) is isomorphic to Hi’1 /Hi .
5. Let E and K be extensions of F , and assume that the composite EK is de¬ned. If A
is any set of generators for K over F (for example, A = K), show that EK = E(A),
the ¬eld formed from E by adjoining the elements of A.
6. Let E/F be a ¬nite Galois extension with Galois group G, and let E /F be a ¬nite
Galois extension with Galois group G . If „ is an isomorphism of E and E with
„ (F ) = F , we expect intuitively that G ∼ G . Prove this formally.
7. Let K/F be a ¬nite separable extension. Although K need not be a normal extension
of F , we can form the normal closure N of K over F , as in (3.5.11). Then N/F
is a Galois extension (see Problem 8 of Section 6.3); let G be its Galois group. Let
H = Gal(N/K), so that the ¬xed ¬eld of H is K. If H is a normal subgroup of G
that is contained in H, show that the ¬xed ¬eld of H is N .
8. Continuing Problem 7, show that H is trivial, and conclude that

gHg ’1 = {1}

where 1 is the identity automorphism.

6.3 Computing a Galois Group Directly
6.3.1 De¬nitions and Comments
Suppose that E is a splitting ¬eld of the separable polynomial f over F . The Galois
group of f is the Galois group of the extension E/F . (The extension is indeed Galois;
see Problem 8.) Given f , how can we determine its Galois group? It is not so easy, but
later we will develop a systematic approach for polynomials of degree 4 or less. Some
cases can be handled directly, and in this section we look at a typical situation. A useful
observation is that the Galois group G of a ¬nite Galois extension E/F acts transitively
on the roots of any irreducible polynomial h ∈ F [X] (assuming that one, hence every,
root of h belongs to E). [Each σ ∈ G permutes the roots by (3.5.1). If ± and β are roots
of h, then by (3.2.3) there is an F -isomorphism of F (±) and F (β) carrying ± to β. This
isomorphism can be extended to an F -automorphism of E by (3.5.2), (3.5.5) and (3.5.6).]

6.3.2 Example
Let d be a positive integer that is√ a perfect cube, and let θ be the positive cube root
not √
= ’ 2 + i 2 3, so that ω 2 = e’i2π/3 = ’ 1 ’ i 1 3 = ’(1 + ω).
1 1
of d. Let ω = e 2 2
The minimal polynomial of θ over the rationals Q is f (X) = X 3 ’ d, because if f were

reducible then it would have a linear factor and d would be a perfect cube. The minimal
polynomial of ω over Q is g(X) = X 2 + X + 1. (If g were reducible, it would have a
rational (hence real) root, so the discriminant would be nonnegative, a contradiction.)
We will compute the Galois group G of the polynomial f (X)g(X), which is the Galois
group of E = Q(θ, ω) over Q.
If the degree of E/Q is the product of the degrees of f and g, we will be able to
make progress. We have [Q(θ) : Q] = 3 and, since ω, a complex number, does not belong
to Q(θ), we have [Q(θ, ω) : Q(θ)] = 2. Thus [Q(θ, ω) : Q] = 6. But the degree of
a ¬nite Galois extension is the size of the Galois group by (3.5.9), so G has exactly 6
automorphisms. Now any σ ∈ G must take θ to one of its conjugates, namely θ, ωθ or
ω 2 θ. Moreover, σ must take ω to a conjugate, namely ω or ω 2 . Since σ is determined by
its action on θ and ω, we have found all 6 members of G. The results can be displayed as
1 : θ ’ θ, ω ’ ω, order = 1
„ : θ ’ θ, ω ’ ω 2 , order = 2
σ : θ ’ ωθ, ω ’ ω, order = 3
σ„ : θ ’ ωθ, ω ’ ω 2 , order = 2
σ 2 : θ ’ ω 2 θ, ω ’ ω, order = 3
„ σ : θ ’ ω 2 θ, ω ’ ω 2 , order = 2
Note that „ σ 2 gives nothing new since „ σ 2 = σ„ . Similarly, σ 2 „ = „ σ. Thus

σ 3 = „ 2 = 1, „ σ„ ’1 = σ ’1 (= σ 2 ). (1)

At this point we have determined the multiplication table of G, but much more insight
is gained by observing that (1) gives a presentation of S3 (Section 5.8, Problem 3). We
conclude that G ∼ S3 . The subgroups of G are

{1}, G, σ , „ , „ σ , „ σ 2

and the corresponding ¬xed ¬elds are

Q, Q(ω), Q(θ), Q(ωθ), Q(ω 2 θ).

To show that the ¬xed ¬eld of „ σ = {1, „ σ} is Q(ωθ), note that „ σ has index 3 in G, so
by the fundamental theorem, the corresponding ¬xed ¬eld has degree 3 over Q. Now „ σ
takes ωθ to ω 2 ω 2 θ = ωθ and [Q(ωθ) : Q] = 3 (because the minimal polynomial of ωθ over
Q is f ). Thus Q(ωθ) is the entire ¬xed ¬eld. The other calculations are similar.

Problems For Section 6.3
1. Suppose that E = F (±) is a ¬nite Galois extension of F , where ± is a root of the
irreducible polynomial f ∈ F [X]. Assume that the roots of f are ±1 = ±, ±2 , . . . , ±n .
Describe, as best you can from the given information, the Galois group of E/F .
2. Let E/Q be a ¬nite Galois extension, and let x1 , . . . , xn be a basis for E over Q.
Describe how you would ¬nd a primitive element, that is, an ± ∈ E such that E =
Q(±). (Your procedure need not be e¬cient.)

3. Let G be the Galois group of a separable irreducible polynomial f of degree n. Show
that G is isomorphic to a transitive subgroup H of Sn . [Transitivity means that if i
and j belong to {1, 2, . . . , n}, then for some σ ∈ H we have σ(i) = j. Equivalently,
the natural action of H on {1, . . . , n}, given by h • x = h(x), is transitive.]
4. Use Problem 3 to determine the Galois group of an irreducible quadratic polynomial
aX 2 + bX + c ∈ F [X], a = 0. Assume that the characteristic of F is not 2, so that
the derivative of f is nonzero and f is separable.
5. Determine the Galois group of (X 2 ’ 2)(X 2 ’ 3) over Q.
6. In the Galois correspondence, suppose that Ki is the ¬xed ¬eld of the subgroup Hi ,
i = 1, 2. Identify the group corresponding to K = K1 © K2 .
7. Continuing Problem 6, identify the ¬xed ¬eld of H1 © H2 .
8. Suppose that E is a splitting ¬eld of a separable polynomial f over F . Show that
E/F is separable. [Since the extension is ¬nite by (3.2.2) and normal by (3.5.7), E/F
is Galois.]
9. Let G be the Galois group of f (X) = X 4 ’ 2 over Q. Thus if θ is the positive fourth
root of 2, then G is the Galois group of Q(θ, i)/Q. Describe all 8 automorphisms in G.
10. Show that G is isomorphic to the dihedral group D8 .
11. De¬ne σ(θ) = iθ, σ(i) = i, „ (θ) = θ, „ (i) = ’i, as in the solution to Problem 10.
Find the ¬xed ¬eld of the normal subgroup N = {1, σ„, σ 2 , σ 3 „ } of G, and verify that
the ¬xed ¬eld is a normal extension of Q.

6.4 Finite Fields
Finite ¬elds can be classi¬ed precisely. We will show that a ¬nite ¬eld must have pn
elements, where p is a prime and n is a positive integer. In addition, there is (up to
isomorphism) only one ¬nite ¬eld with pn elements. We sometimes use the notation
GF (pn ) for this ¬eld; GF stands for “Galois ¬eld”. Also, the ¬eld with p elements will
be denoted by Fp rather than Zp , to emphasize that we are working with ¬elds.

6.4.1 Proposition
Let E be a ¬nite ¬eld of characteristic p. Then |E| = pn for some positive integer n.
Moreover, E is a splitting ¬eld for the separable polynomial f (X) = X p ’ X over Fp , so
that any ¬nite ¬eld with pn elements is isomorphic to E. Not only is E generated by the
roots of f , but in fact E coincides with the set of roots of f .
Proof. Since E contains a copy of Fp (see (2.1.3), Example 2), we may view E as a vector
space over Fp . If the dimension of this vector space is n, then since each coe¬cient in a
linear combination of basis vectors can be chosen in p ways, we have |E| = pn .
Now let E — be the multiplicative group of nonzero elements of E. If ± ∈ E — , then
±p ’1 = 1 by Lagrange™s theorem, so ±p = ± for every ± ∈ E, including ± = 0. Thus
n n

each element of E is a root of f , and f is separable by (3.4.5). Now f has at most pn
distinct roots, and as we have already identi¬ed the pn elements of E as roots of f , in
fact f has pn distinct roots and every root of f must belong to E. ™


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