## Every finite group has a composition series

Prove that every finite group has a composition series.

Let $G$ be a finite group.

If $|G| = 1$, then $G \cong 1$, so that $G$ has a trivial composition series.

Suppose that every group of order less than or equal to $n$ has a composition series, and let $G$ be a group of order $n+1$. If $G$ is simple, then $1 \vartriangleleft G$ is a composition series for $G$. If $G$ is not simple, then $G$ has a nontrivial normal subgroup $N$. Now $|N| \leq n$ and $|G/N| \leq n$, so that both $N$ and $G/N$ have composition series; say $1 = H_1 \vartriangleleft H_2 \vartriangleleft \cdots \vartriangleleft H_{k-1} \vartriangleleft H_k = N$ and $1 = K_1/N \vartriangleleft K_2/N \vartriangleleft \cdots \vartriangleleft K_{\ell-1}/N \vartriangleleft K_\ell/N = G/N$. Then $1 = H_1 \vartriangleleft H_2 \vartriangleleft \cdots \vartriangle H_k = K_1 \vartriangleleft \cdots \vartriangleleft K_{\ell-1} \vartriangleleft K_\ell = G$ is a composition series for $G$. By induction, every finite group has a composition series.