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?