The following is an exercise from an algebra course I am taking:
Let $G$ be a simple group with composition series $G = G_0 \vartriangleright G_1 \vartriangleright \cdots \vartriangleright G_n = \{1\}$. Let $N$ be a normal subgroup of $G$. Show that if $N$ is simple, then there exists exactly one index $m$ such that $G_m N = G_{m + 1} N$.
So far I've been able to show that there exists another composition series $G = H_0 \vartriangleright \cdots \vartriangleright H_n = \{1\}$ of the same length as the original with $H_{n - 1} = N$ and so $H_{n - 1} N = H_n N = N$. Then by Jordan Holder, we have that $H_{n - 1} / H_n$ is isomorphic to some $G_m / G_{m - 1}$, and both are isomorphic to $N$. After that, I'm not really sure where to go. Any pointers?