Difference between revisions of "Jordan-Hölder Theorem"

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The '''Jordan-Hölder Theorem''' is a result in [[group theory]], named for Camille Jordan and Otto Hölder.  It states that any two [[Jordan-Hölder series]] of the same group are equivalent.  Jordan proved that the cardinalities of the quotients are invariant up to order in 1869 (?); Hölder proved that the quotients are in fact isomorphic in 1889.  In 1928, Schreier published an improved proof using [[Schreier's Theorem]].  Six years later, in 1934, Zassenhaus improved Scheier's proof, using [[Zassenhaus's Lemma]].
 
The '''Jordan-Hölder Theorem''' is a result in [[group theory]], named for Camille Jordan and Otto Hölder.  It states that any two [[Jordan-Hölder series]] of the same group are equivalent.  Jordan proved that the cardinalities of the quotients are invariant up to order in 1869 (?); Hölder proved that the quotients are in fact isomorphic in 1889.  In 1928, Schreier published an improved proof using [[Schreier's Theorem]].  Six years later, in 1934, Zassenhaus improved Scheier's proof, using [[Zassenhaus's Lemma]].
  

Revision as of 21:53, 19 May 2008

This is an AoPSWiki Word of the Week for May 20-27

The Jordan-Hölder Theorem is a result in group theory, named for Camille Jordan and Otto Hölder. It states that any two Jordan-Hölder series of the same group are equivalent. Jordan proved that the cardinalities of the quotients are invariant up to order in 1869 (?); Hölder proved that the quotients are in fact isomorphic in 1889. In 1928, Schreier published an improved proof using Schreier's Theorem. Six years later, in 1934, Zassenhaus improved Scheier's proof, using Zassenhaus's Lemma.

Proof

Suppose $\Sigma_1$ and $\Sigma_2$ are two Jordan-Hölder series for the same group. By Schreier's Theorem, there exist equivalent composition series $\Sigma'_1$ and $\Sigma'_2$, finer than $\Sigma_1$ and $\Sigma_2$ respectively. But since $\Sigma_1$ and $\Sigma_2$ are Jordan-Hölder composition series, they are obtained from $\Sigma'_1$ and $\Sigma'_2$ by removing repetitions; hence they are equivalent. $\blacksquare$

See also