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

$$G\supset G_1\supset G_2\supset \mathrm{\cdots }\supset G_n=\{1\}$$

and

$$G\supset G_1^{\prime }\supset G_2^{\prime }\supset \mathrm{\cdots }\supset G_m^{\prime }=\{1\}$$

are two decompositions of $G$. If $G_1=G_1^{\prime }$ then we’re done (apply the induction hypothesis to $G_1$), so assume $G_1\ne G_1^{\prime }$. Set $H:=G_1\cap G_1^{\prime }$ and choose a decomposition series

$$H\supset H_1\supset \mathrm{\cdots }\supset H_k=\{1\}$$

for $H$. By the second isomorphism theorem, $G_1/H=G_1{G}_1^{\prime }/G_1^{\prime }=G/G_1^{\prime }$ (the last equality is because $G_1{G}_1^{\prime }$ 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

$$G_1\supset G_2\supset \mathrm{\cdots }\supset G_n$$

and

$$G_1\supset H\supset \mathrm{\cdots }\supset H_k$$

are two decomposition series for $G_1$, and hence have the same simple quotients by the induction hypothesis; likewise for the $G_1^{\prime }$ series. Therefore $n=m$. Moreover, since $G/G_1=G_1^{\prime }/H$ and $G/G_1^{\prime }=G_1/H$ (by the second isomorphism theorem), we have now accounted for all of the simple quotients, and shown that they are the same.