Question

Jordan holder theorem Abstract algebra​

Answer 1

Let G be a finite group. Suppose we have two chains of subgroups
0 = G0 ( G1 ( G2 ( · · · ( Gr = G
and
0 = H0 ( H1 ( H2 ( · · · ( Hs = G
so that Gi
is normal in Gi+1, with Gi+1/Gi simple and Hj is normal in Hj+1, with Hj+1/Hj simple.
Theorem (Jordan-Holder): In the above setting, we have r = s, and the list of quotients
(G1/G0, G2/G1, . . . , Gr/Gr−1) is a rearrangement of the list of quotients (H1/H0, H2/H1, . . . , Hs/Hs−1).
The set up of the theorem is easy to achieve:
Theorem 2: Let G be any finite group. Then we can find a chain of normal subgroups
0 = G0 ( G1 ( G2 ( · · · ( Gr = G
so that Gi
is normal in Gi+1 and Gi+1/Gi
is simple.
Proof: Let r be the largest possible number so that there is a chain
0 = G0 ( G1 ( G2 ( · · · ( Gr = G
with Gi
is normal in Gi+1. We can always take r = 1, and (G0, G1) = ({1}, G), so such an r exists
and, since G is finite, there is a largest such r.
We claim that Gi+1/Gi
is simple. If not, let H be a normal subgroup of Gi/Gi+1 and let π be
the projection map Gi → Gi/Gi+1. Then π
−1
(H) is normal in Gi+1, and Gi
is normal in H. So
0 = G0 ( G1 ( G2 ( · · · ( Gi ( π
−1
(H) ( Gi+1 ( · · · ( Gr = G
is a longer chain, contradiction.
Such a chain is called a composition series for G.
Preliminary lemmas
Lemma 1: Let G be a group, H a normal subgroup and A an arbitrary subgroup of G. Then
AH is a subgroup of G, where AH is the set of products {ah : a ∈ A, h ∈ H}.
Proof 1: We need to show that AH is closed under multiplication and inverses. For the first,
let a1, a2 ∈ A and h1, h2 ∈ H. Notice that
(a1h1)(a2h2) = (a1a2)(a
−1
2 h1a2h2).
Since H is normal, we have a
−1
2 h1a2 ∈ H. So (a
−1
2 h1a2h2) ∈ H and, clearly, a1a2 ∈ A.
Similarly, for a ∈ A and h ∈ H, we have
(ah)
−1 = h
−1
a
−1 = a
−1
(ah−1
a
−1
)
and we see that (ah)
−1 ∈ AH.
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