Jordan -Holder for infinite series? composition series $\rightarrow$ chief series The problem I am trying  to prove is that a group which has a composition series will also have a chief series (we 're only talking about infinite groups of course). Instead of just posting the answer I would appreciate if someone told me if my line of thought is correct as I am posting several other questions there.
I have reduced the problem to just showing that there is a minimal normal subgroup. Let $G>G_1>G_2>....G_r=1$ be a composition series. My thought here is to use contradiction and say there in no minimal normal subgroup. We have that $G_1\unlhd G$ and thus there is a subgroup $N<G_1$ such that $N\unlhd G$. This way I can construct an infinite normal series which does not terminate in which $N_{i+1}\unlhd G$ and $N_{i+1}\unlhd N_i$. I could also take the first r of the $N_i$ and claim that $N_r$ has a composition series. All I need in the end is to refine that series to start from $G$ and then use Jordan Holder theorem. 
What I am confused about is whether I can use the Jordan Holder Theorem which states any composition series have the same length and they are equivalent (all I need is the length part). Can I use the Jordan Holder theorem to claim that any normal series(need not be composition) is of length $\leq r$?. The subgroups $N_i$ do construct a normal series but I cannot seem to find a way to refine it into a composition series. What I mean is this: Take the first $N_0=N$. We have that $N\leq G_1$ but we do not have that this is the maximal normal subgroup. If such a group existed then we would be finished, constructing the composition series as planned. But how can we show that there is a maximal normal subgroup between $G_1$ and $N$, e.g $N\leq H \leq G_1$ in which $H$ is maximal in $G_1$?
I think my question can be reduced to this problem: If $G$ has a maximal normal  subgroup $K$ and $N\unlhd G$, then is $N$ contained in some maximal normal subgroup? If $N\leq K$ the answer is obvious, but if that is not the case, can we still argue that $N$ is contained in some maximal normal subgroup? My only thought was to take $NK=G$ but i am confused on how to continue.
My questions may  not even be linked in the end to each other but any help would be appreciated!
 A: It is a bit hard to figure out exactly what you are asking. Let's prove the following result. If $G$ has a composition series of length $r$, then any strictly descending subnormal series of $G$ has length at most $r$. Will that answer your question?
We can prove that by induction on $r$. If $r=1$ then $G$ is simple and the result holds.
Otherwise let $N$ be a normal subgroup of $G$ with $N \ne G$. It will be enough to prove that any strictly descending subnormal series of $N$ has length at most $r-1$.
If $N \le G_1$ then this follows immediately by induction applied to $G_1$. So suppose not, and hence $G_1N=G$.
Then $G_1 > G_1 \cap N \ge G_2 \cap N \ge \cdots \ge G_r \cap N = 1$ is a subnormal series of $G_1$. Since $G_1$ has a composition series of length $r-1$, by induction  we must have $G_i \cap N = G_{i+1} \cap N$ for some $i \ge 1$, or else we would have a strictly subnormal descending series of $G_1$ of length greater than $r-1$.
Now consider the subnormal series $N >G_1 \cap  N \ge G_2 \cap N \ge \cdots \ge G_r \cap N = 1$ of $N$. Its factors are isomorphic to normal subgroups of the simple groups $G_i/G_{i+1}$, so they are either simple or trivial. But by the previous paragraph at least one of these factors is trivial, so the series refines to a composition series of $N$ of length at most $r-1$. Now the result follows by induction applied to $N$.
