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theorem says that they all have the same length, and the same quotients (up to order and
isomorphism). More precisely:
Theorem 6.2 (Jordan-Hölder). If
G = G0 G1 · · · Gs = {1}
G = H0 H1 · · · Ht = {1}
are two composition series for G, then s = t and there is a permutation À of {1, 2, . . . , s}
13
such that Gi/Gi+1 H" HÀ(i)/HÀ(i+1).
Proof. We use induction on the order of G.
Case I: H1 = G1. In this case, we have two composition series for G1, to whic h we c an
apply the induction hypothesis.
Case II: H1 = G1. Because each of G1 and H1 is normal in G, G1H1 is a normal subgroup
of G, and it properly contains both G1 and H1. But they are maximal normal subgroups of
G, and so G1H1 = G. Therefore
G/G1 = G1H1/G1 H" H1/G1 )" H1 (see 3.2).
Similarly G/H1 H" G1/G1 )" H1. Hence K2 =df G1 )" H1 is a maximal normal subgroup in
both G1 and H1, and
G/G1 H" H1/K2, G/H1 H" G1/K2.
Choose a composition series
K2 K3 · · · Ku.
We have the picture:
G1 G2 · · · Gs
GK2 · · · Ku .
H1 H2 · · · Ht
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Jordan showed that corresponding quotients had the same order, and Hölder that they were isomorphic.
48 J.S. MILNE
On applying the induction hypothesis to G1 and H1 and their composition series in the
diagram, we find that
Quotients(G G1 G2 · · · )
In passing from the second to the third line, we used the isomorphisms G/G1 H" H1/K2 and
G/H1 H" G1/K2.
Note that the theorem applied to a cyclic group Cm implies that the factorization of an
integer into a product of primes is unique.
Remark 6.3. There are infinite groups having finite composition series (there are infinite
simple groups). For such a group, let d(G) be the minimum length of a composition series.
Then the Jordan-Hölder theorem extends to show that all composition series have length
d(G) and have isomorphic quotient groups. The same proof works: use induction on d(G)
instead of (G : 1).
The quotients of a composition series are also called composition factors. (Some authors
call a quotient group G/N a factor group of G; I prefer to reserve this term for a subgroup
H of G such that G = H × H .)
6.2. Solvable groups.
Agroupis solvable if it has a normal series whose quotient groups are all commutative. Such
a series is called a solvable series. Alternatively, we can say that a group is solvable if it can
be obtained by forming successive extensions of abelian groups. Since a commutative group
is simple if and only if it is cyclic of prime order, we see that G is solvable if and only if for
one (hence every) composition series the quotients are all cyclic groups of prime order.
Any commutative group is solvable, as is any dihedral group. The results in Section 5
show that every group of order
group, e.g., An for n e" 5, will not be solvable.
There is the following result:
Theorem 6.4 (Feit-Thompson 1963). Every finite group of odd order is solvable.
Proof. The proof occupies a whole issue of the Pacific J. Math., and hence is omitted.
This theorem played a very important role in the development of group theory, because it
shows that every noncommutative finite simple group contains an element of order 2. It was
a starting point in the program that eventually led to the classification of all finite simple
groups.
" " 1 "
Example 6.5. Consider the subgroups G = and G1 = of GL2(k),
0 " 0 1
some field k. Then G1 is a normal subgroup of G, and G/G1 H" k× × k×, G1 H" (k, +). Hence
G is solvable.
GROUP THEORY 49
Proposition 6.6. (a) Every subgroup and every quotient group of a solvable group is solv-
able.
(b) An extension of solvable groups is solvable.
Proof. (a) Let G G1 · · · Gn be a solvable series for G, and let H be a subgroup of G.
The homomorphism
x ’! xGi+1 : H )" Gi ’! Gi/Gi+1
has kernel (H )" Gi) )" Gi+1 = H )" Gi+1. Therefore H )" Gi+1 is a normal subgroup of H )" Gi
and the quotient H )" Gi/H )" Gi+1 injects into Gi/Gi+1, and so is commutative. We have
shown that
H H )" G1 · · · H )" Gn
is a solvable series for H.
Let be a quotient group of G, and let i be the image of Gi in . Then
1 · · · n = {1}
is a solvable series for .
(b) Let N be a normal subgroup of G, and let = G/N. We have to show that if N and
are solvable, then so also is G. Let
1 · · · n = {1}
N N1 · · · Nm = {1}
be a solvable series for and N, andlet Gi be the inverse image of i in G. Then Gi/Gi+1 H"
i/ i+1 (see 3.4), and so the
G G1 · · · Gn(= N) N1 · · · Nm
is a solvable series for G.
Corollary 6.7. A finite p-group is solvable.
Proof. We use induction on the order the group G. According to (4.14), the centre Z(G)
of G is nontrivial, and so the induction hypothesis shows that G/Z(G) is solvable. Because
Z(G) is solvable, Proposition 6.6b shows that G is solvable.
Let G be a group. Recall that the commutator of x, y " G is
[x, y] =xyx-1y-1 = xy(yx)-1
Thus [x, y] =1 Ð!Ò! xy = yx, and G is commutative Ð!Ò! all commutators are 1.
Example 6.8. For any finite-dimensional vector space V over a field k and any full flag
F = {Vn, Vn-1, . . . } in V , the group
B(F ) ={± " Aut(V ) | ±(Vi) ‚" Vi all i}
is solvable. When k = Fp, this can be proved by noting that B(F )/P (F ) is commutative,
and that P (F ) is a p-group and is therefore solvable. The general case is left as an exercise.
50 J.S. MILNE
For any homomorphism Õ : G ’! H
Õ[x, y] =Õ(xyx-1y-1) =[Õ(x), Õ(y)]
i.e., Õ maps the commutator of x, y to the commutator of Õ(x), Õ(y). In particular, we see
that if H is commutative, then Õ maps all commutators in G to 1.
The group G generated by the commutators in G is called the commutator or first derived
subgroup of G.
Proposition 6.9. The commutator subgroup G is a characteristic subgroup of G; it is the
smallest normal subgroup of G such that G/G is commutative.
Proof. An automorphism ± of G maps the generating set for G into G , and hence maps G
into G . Since this is true for all automorphisms of G, G is characteristic.
¯
Write g ’! ! for the map g ’! gG : G ’! G/G ; then [g, h] ’! [! , h]; but [g, h] ’! 1 and so
¯ ¯
[! , h] = 1 for all ! , h " G/G . Hence G/G is commutative.
If N is normal and G/N is commutative, then [g, h] ’! 1 in G/N, and so [g, h] " N. Sinc e
these elements generate G , N ƒ" G .
For n e" 5, An is the smallest normal subgroup of Sn giving a commutative quotient.
Hence (Sn) = An.
The second derived subgroup of G is (G ) ; the third is G(3) = (G ) ; and so on. Each
derived group is a characteristic subgroup of G. Hence we obtain a normal series
G ƒ" G ƒ" G(2) ƒ" · · · ,
which is called the derived series. For example, if n e" 5, then the derived series of Sn is
Sn ƒ" An ƒ" An ƒ" An ƒ" · · · .
Proposition 6.10. A group G is solvable if and only if its kth derived subgroup G(k) =1
for some k.
Proof. If G(k) = 1, then the derived series is a solvable series for G. Conversely, let
G = G0 G1 G2 · · · Gs = {0}
be a solvable series for G. Bec ause G/G1 is commutative, G1 ƒ" G . Now G G2 is a subgroup
of G1, and from
H"
G /G )" G2 ’! G G2/G2 ‚" G1/G2
we see that
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