proof of the Jordan Hölder decomposition theorem
Let . We first prove existence, using induction on . If (or, more generally, if is simple) the result is clear. Now suppose is not simple. Choose a maximal proper normal subgroup of . Then has a Jordan–Hölder decomposition by induction, which produces a Jordan–Hölder decomposition for .
To prove uniqueness, we use induction on the length of the decomposition series. If then is simple and we are done. For , suppose that
and
are two decompositions of . If then we’re done (apply the induction hypothesis to ), so assume . Set 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 |
---|---|
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 |