# 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 $G_{1}$ of $G$. Then $G_{1}$ 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

 $G\supset G_{1}\supset G_{2}\supset\cdots\supset G_{n}=\{1\}$

and

 $G\supset G^{\prime}_{1}\supset G^{\prime}_{2}\supset\cdots\supset G^{\prime}_{% m}=\{1\}$

are two decompositions of $G$. If $G_{1}=G^{\prime}_{1}$ then we’re done (apply the induction hypothesis to $G_{1}$), so assume $G_{1}\neq G^{\prime}_{1}$. Set $H:=G_{1}\cap G^{\prime}_{1}$ and choose a decomposition series

 $H\supset H_{1}\supset\cdots\supset H_{k}=\{1\}$

for $H$. By the second isomorphism theorem, $G_{1}/H=G_{1}G^{\prime}_{1}/G^{\prime}_{1}=G/G^{\prime}_{1}$ (the last equality is because $G_{1}G^{\prime}_{1}$ is a normal subgroup of $G$ properly containing $G_{1}$). In particular, $H$ is a normal subgroup of $G_{1}$ with simple quotient. But then

 $G_{1}\supset G_{2}\supset\cdots\supset G_{n}$

and

 $G_{1}\supset H\supset\cdots\supset H_{k}$

are two decomposition series for $G_{1}$, and hence have the same simple quotients by the induction hypothesis; likewise for the $G^{\prime}_{1}$ series. Therefore $n=m$. Moreover, since $G/G_{1}=G^{\prime}_{1}/H$ and $G/G^{\prime}_{1}=G_{1}/H$ (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 ProofOfTheJordanHolderDecompositionTheorem 2013-03-22 12:08:49 2013-03-22 12:08:49 djao (24) djao (24) 9 djao (24) Proof msc 20E22