Assume we have a composition series $\{e\} = G_0 \triangleleft G_1 \triangleleft \dots \triangleleft G_k = G$, i.e. each $G_{i+1}/G_i$ is simple. Now we take some normal subgroup $N$ of $G$. We can create a new (non-strict) composition series as follows:
$$ \{e\} = G_0 \cap N \trianglelefteq G_1 \cap N \trianglelefteq \dots \trianglelefteq G_k \cap N = N = G_0 N \trianglelefteq G_1 N \trianglelefteq \dots \trianglelefteq G_k N = G. $$
We can then look at $(G_{i+1} \cap N) / (G_i \cap N)$ and $(G_{i+1} N) / (G_i N)$ to determine at which steps the inclusion is strict and at which it is not.
I just didn't fully understand one detail:
- Why is $G_i \cap N$ normal in $G_{i+1} \cap N$?
- Analogously, why is $G_i N$ normal in $G_{i+1} N$?
I think number 1 is easy: Because $G_{i}$ is normal in $G_{i+1}$, we have $\pi: G_{i+1} \to G_{i+1} / G_i$ with $\ker(\pi) = G_i$. We can then just construct $$ f: G_{i+1} \cap N \to G_{i+1} / G_i, g \mapsto \pi(g) $$ and $\ker(f) = \ker(\pi) \cap N = G_i \cap N$, therefore $G_i \cap N$ is normal in $G_{i+1} \cap N$.
Now I would like to the the same for 2), i.e. construct a homomorphism $f: G_{i+1}N \to X$ with $\ker(f) = G_i N$, but I'm not sure how to construct one. I'd like to reuse $\pi$, but extending it to $h: G_{i+1} N \to G_{i+1}/G_i$ doesn't seem to work, as representing $x \in G_{i+1}N$ via $x = g \cdot n, g \in G_{i+1}, n \in N$ isn't unique. (or is it?)
I'm stuck here, and help would be appreciated.
$G_{i+1} \le N_G(G_i)$ by assumption, and $G_{i+1} \le N_G(N)$ because $N$ is a normal subgroup of $G$. So $G_{i+1} \le N_G(G_{i}N)$.
Also $N \le N_G(G_{i}N)$ because $ N \le G_{i}N$.
So $G_{i+1}N \le N_G(G_iN)$ and hence $G_iN$ is normal in $G_{i+1}N$.