The Jordan–Hölder Theorem

Proof: Let G be a finite group. We know at least one subnormal series,
namely G 1. Let
G = G 0 >G 1 > ···>G n = 1 (4.2)
be a longest possible subnormal series of G. ThiscertainlyexistssinceG has
only finitely many subgroups, so only finitely many subnormal series. We
claim that (4.2) is a composition series.
Suppose that it is not. Then one of the factors, say G j /G j+1 ,isnot
simple. This quotient then possesses a non-trivial proper normal subgroup
and this corresponds to a subgroup N of G with
Then
G j+1 G 1 > ···>G j >N>G j+1 > ···>G n = 1
is a subnormal series in G (note that G j+1 N since G j+1 G j )whichis
longer than (4.2). This contradicts our assumption that (4.2) is the longest
such series. Hence (4.2) is indeed a composition series for G.
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On the other hand, we have an example of an infinite group (namely the
infinite cyclic group) which does not possess a composition series. There are
infinite groups that possess composition series, but infinite simplegroups
(which necessarily occur as some of the composition factors) aremuchless
well understood than finite simple groups.
The important thing about composition series is that the composition
factors occurring are essentially unique. This is the content of the following
important theorem.
Theorem 4.6 (Jordan–Hölder Theorem) Let G be a group and let
and
G = G 0 >G 1 >G 2 > ··· >G n = 1
G = H 0 >H 1 >H 2 > ···>H m = 1
be composition series for G. Thenn = m and there is a one-one correspondence
between the two sets of composition factors
and
{G 0 /G 1 ,G 1 /G 2 ,...,G n−1 /G n }
{H 0 /H 1 ,H 1 /H 2 ,...,H m−1 /H m }
such that corresponding factors are isomorphic.
Proof: See Tutorial Sheet IV.
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