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!

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 $N$.

• First of all thanks for the answer. However, I am a bit confused with the proof. I had thought of taking $G_i\cap N$ myself but it seems to me that you're only using a specific subnormal series of N, namely the series $N>G_1\cap N \geq G_1\cap N \geq \cdots \geq G_r \cap N=1$. How can you generalize it after that for any strictly descending subnormal series of N? I suppose it's the step that you say in the last sentece "the rest follows by induction on N", but could you please elaborate a little on that? – Foivos Apr 7 '18 at 9:24
• We have proved that $N$ has a composition series of length at most $r-1$. So that means that we can apply our inductive hypothesis to $N$ to say that any strictly descending subnormal series of $N$ has length at most $r-1$. – Derek Holt Apr 7 '18 at 11:08