Sheng Li (gmachine1729) wrote,
Sheng Li
gmachine1729

Composition series

Originally published at 狗和留美者不得入内. You can comment here or there.

My friend after some time in industry is back in school, currently taking graduate algebra. I was today looking at one of his homework and in particular, I thought about and worked out one of the problems, which is to prove the uniqueness part of the Jordan-Hölder theorem. Formally, if G is a finite group and

1 = N_0 \trianglelefteq N_1 \trianglelefteq \cdots \trianglelefteq N_r = G and 1 = N_0' \trianglelefteq N_1' \trianglelefteq \cdots \trianglelefteq N_s' = G

are composition series of G, then r = s and there exists \sigma \in S_r and isomorphisms N_{i+1} / N_i \cong N_{\sigma(i)+1} / N_{\sigma(i)}.

Suppose WLOG that s \geq r and as a base case s = 2. Then clearly, s = r and if N_1 \neq N_1', N_1 \cap N_1' = 1. N_1 N_1' = G must hold as it is normal in G. Now, remember there is a theorem which states that if H, K are normal subgroups of G = HK with H \cap K = 1, then G \cong H \times K. (This follows from (hkh^{-1})k^{-1} = h(kh^{-1}k^{-1}), which shows the commutator to be the identity). Thus there are no other normal proper subgroups other than H and K.

For the inductive step, take H = N_{r-1} \cap N_{s-1}'. By the second isomorphism theorem, N_{r-1} / H \cong G / N_{s-1}'. Take any composition series for H to construct another for G via N_{r-1}. This shows on application of the inductive hypothesis that r = s. One can do the same for N_{s-1}'. With both our composition series linked to two intermediary ones that differ only between G and the common H with factors swapped in between those two, our induction proof completes.

Tags: algebra, group theory, 数学/математика
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