If $H_i \unlhd H_{i+1}$ and $N$ is simple then $H_iN \unlhd H_{i+1} N$

Question

Let $G$ be a group, $\{H_i\} $ a composition series and $N \le G$ simple. Show that $\{H_iN\} $ is also a composition series.

Actually, the question in the title is a hint to prove that statement, but I haven't managed to get it right.

My first attempt was to reorder the factors in $(h_{i+1}n)(h_in')(n^{-1}h_{i+1}^{-1}) $, but we can't even assume $hn_1 = n_2h$ since the $H$ subgroups might not be normal in $G$. I also tried inserting $(h_{i+1}nh_in^{-1}h_{i+1}^{-1}) (h_{i+1}nn'n^{-1}h_{i+1}^{-1})$, but I can't see any path to the solution from here.

Any hints on how to proceed? I feel this may be rather trivial but I'm just stuck and can't come up with anything useful. I know this is probably because I'm not using anywhere the fact that $N$ and $H_{i+1}/H_i$ are simple, but I don't know the implications this has.

Thanks for your help.

Answer 1

With

$$G=S_3\;,\;\;H_1:=(1)\lhd H_2:=A_3\lhd H_3:=S_3\;,\;\;N=\langle (12)\rangle$$

we get:

$$H_1N=N\;,\;\;H_2N=S_3\;,\;\;\text{yet}\;\;H_1N=N\rlap{\;\;/}\lhd H_2N=S_3$$

and we get a counterexample...or I missed something. Check this.