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Abstract Algebra
Notes xyx y = a baba bab, and so we get
-1 -1
j
j
i
i
xyx y = a a b a a b = a 2(i-j) , and again we get even powers of a.
-1 -1
i -j 2 i -j 2
Thus the commutator subgroup D is either < a > (if n is odd) or < a > (if n is even). In either case,
2
n
the commutator subgroup is abelian, so D = {e}.
n
Example: Prove that any group of order 588 is solvable, given that any group of order 12
is solvable.
We have 588 = 2 · 3 · 7 . Let S be the Sylow 7-subgroup. It must be normal, since 1 is the only
2
2
divisor of 12 that is 1 mod 7. By assumption, G / S is solvable since | G / S | = 12. Furthermore,
S is solvable since it is a p-group. Since both S and G / S are solvable, it follows from Corollary
that G is solvable.
Example: Let G be a group of order 780=2 · 3 · 5 · 13. Assume that G is not solvable. What
2
are the composition factors of G? (Assume that the only nonabelian simple group of order 60 is
A .)
5
The Sylow 13-subgroup N is normal, since 1 is the only divisor of 60 that is 1 mod 13. Using the
fact that the smallest simple nonabelian group has order 60, we see that the factor G/N must be
simple, since otherwise each composition factor would be abelian and G would be solvable.
Thus the composition factors are Z and A . 5
13
Theorem-[Jordan-Hölder] Any two composition series for a finite group have the same length.
Furthermore, there exists a one-to-one correspondence between composition factors of the two
composition series under which corresponding composition factors are isomorphic.
Let |G| = N. We first prove existence, using induction on N. If N = 1 (or, more generally, if G is
simple) the result is clear. Now suppose G is not simple. Choose a maximal proper normal
subgroup G1 of G. Then G1 has a Jordan-Hölder decomposition by induction, which produces a
Jordan-Hölder decomposition for G.
To prove uniqueness, we use induction on the length n of the decomposition series. If n=1 then
G is simple and we are done. For n > 1, suppose that
G G1 G2 Gn = {1}
and
G G1 G2 Gm=1
are two decompositions of G . If G1 = G1 then were done (apply the induction hypothesis to G1),
so assume G1/G1 . Set H : = G1 G1 and choose a decomposition series H H Hk = {1} for H.
1
By the second isomorphism theorem, G1/H=G1G1/G1=G/G1 (the last equality is because G1G1
is a normal subgroup of G properly containing G1). In particular, H is a normal subgroup of G1
with simple quotient. But then
G1 G2 ... Gn
and
G1 H ... Hk
are two decomposition series for G1, and hence have the same simple quotients by the induction
hypothesis; likewise for the G1 series. Therefore, n=m. Moreover, since G/G1=G1/H and
G/G1=G1/H (by the second isomorphism theorem), we have now accounted for all of the
simple quotients, and shown that they are the same.
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