proof of the Jordan Hölder decomposition theorem
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⊃G′1⊃G′2⊃⋯⊃G′m={1} |
are two decompositions of G. If G1=G′1 then we’re done (apply the induction hypothesis to G1), so assume G1≠G′1. Set H:= and choose a decomposition series
for . By the second isomorphism theorem, (the last equality is because is a normal subgroup of properly containing ). In particular, is a normal subgroup of with simple quotient
. But then
and
are two decomposition series for , and hence have the same simple quotients by the induction hypothesis; likewise for the series. Therefore . Moreover, since and (by the second isomorphism theorem), we have now accounted for all of the simple quotients, and shown that they are the same.
Title | proof of the Jordan Hölder decomposition theorem![]() |
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Canonical name | ProofOfTheJordanHolderDecompositionTheorem |
Date of creation | 2013-03-22 12:08:49 |
Last modified on | 2013-03-22 12:08:49 |
Owner | djao (24) |
Last modified by | djao (24) |
Numerical id | 9 |
Author | djao (24) |
Entry type | Proof |
Classification | msc 20E22 |