I want to show that if $N$ is a simple normal subgroup of a group $G$ such that $G/N$ has a composition series, then also $G$ has a composition series. I think I can finish the proof if I can only show the following statement:
If $N \leq G_1 \leq G$ and $N \unlhd G$ are such that $G_1/N \unlhd G/N$, then also $G_1 \unlhd G$.
However, I cannot think of a way to prove this and I'm unsure if this statement is even true.
It seems to me that the straightforward way of proving normality of a subgroup, i.e. showing that $\forall g \in G$ we have $gG_1 = G_1g$, is not the best way in this case. So I tried to find an appropriate homomorphism from $G$ to some other group that has kernel $G_1$ but I couldn't think of anything.
Can someone help?
Actually this is a very basic exercise in group theory. Consider this composition of projections: $$G \longrightarrow G/N \longrightarrow (G/N)/(G_1/N)$$ The kernel is $G_1$, so $G_1$ is normal. Moreover this show that $$(G/N)/(G_1/N) \cong G/G_1$$