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I'm having trouble understanding one step in the proof that every finite group has a composition series. We proved this with induction over $d = |G|$ in our algebra lectures. The case $d=1$ is trivial. For $d > 1$, let $H$ be a real normal subgroup of $G$ with minimal index $[G:H]$. Now we claim that $G/H$ is simple (i.e. has no normal subgroups apart from itself and the trivial subgroup). Why is this true?

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If $G/H$ is not simple, then there exists a subgroup of $G$ with index smaller than that of $H$, contradiction.

In particular let $K/H$ be a proper subgroup (correspondence theorem) of $G/H$ that is not zero. Then $|G:H|=|G:K| |K:H| \Rightarrow |G:K| < |G:H|$.

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