The Jordan–Hölder Theorem

is a composition series, then the composition factors
G 0 /G 1 ,G 1 /G 2 ,...,G m−1 /G m
are abelian simple groups. They are therefore cyclic of prime order. Now
|G| = |G 0 /G 1 |·|G 1 /G 2 |·...·|G m−1 /G m |.
This must be the prime factorisation of |G| = n, andhencethecomposition
factors of G are
C p1 ,C p1 ,...,C p1 ,C p2 ,C p2 ,...,C p2 ,...,C ps ,C ps ,...,C ps .
} {{ } } {{ } } {{ }
r 1 times
r 2 times
r s times
Although the Jordan–Hölder Theorem tells us that the composition factors
are essentially uniquely determined, the composition series need not be
unique. For example, if G = 〈x〉 is cyclic of order 30, then there are several
different composition series; e.g.,
G = 〈x〉 > 〈x 2 〉 > 〈x 6 〉 > 1
where the composition factors are C 2 , C 3 and C 5 ,and
G = 〈x〉 > 〈x 3 〉 > 〈x 15 〉 > 1
where the composition factors are C 3 , C 5 and C 2 ;etc.
Later in the course we shall characterise the finite groups whose composition
factors are cyclic as being the soluble groups.
Our final example has a unique composition series:
Example 4.10 Let n 5andconsiderthesymmetricgroupS n of degree n.
We already know the following series:
S n >A n > 1 (4.3)
which has factors C 2 and A n . Both of these are simple groups, so (4.3) is
acompositionseriesforS n . It can be shown that S n has precisely three
normal subgroups (namely those occurring in the above series) and hence
(4.3) is the only composition series for S n .
The Jordan–Hölder Theorem again raises the question of how we put the
composition factors back together. We have a unique decomposition, but
how complicated is the reverse process The answer turns out to be rather
difficult, but in the next chapter we shall meet some ways of creating new
groups and this will give some ways of putting the composition factorsback
together.
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