Let $G$ be a finite group and $L$ a maximal subgroup of $G$, then all minimal normal subgroups $N$ of $G$ that satisfy $N\cap L = 1$ are isomorphic.

My attempt:

By the Second Isomorphism Theorem, we have $LN/N\cong L/(N\cap L)=L$, where $LN$ is precisely $G$ since $N$ is normal and $L$ is maximal.

Therefore for all minimal normal subgroups $N_1, N_2$ of $G$, $G/N_1\cong L\cong G/N_2$. But I’m stuck here, does $G/N_1\cong G/N_2$ imply $N_1\cong N_2$? I believe that the minimality of $N$ as a normal subgroup must be the key to my question, but how to do it?

Any help is sincerely appreciated! Thanks!

PS: It’s exercise 1 on page 39 of my textbook, The Theory of Finite Groups, An Introduction.


In general, $G/N_1\cong G/N_2$ does not imply $N_1\cong N_2$, but (for finite groups) it does imply that $N_1$ and $N_2$ have the same multiset of composition factors. Now, in this particular case, you know that $N_1$ and $N_2$ are direct powers of simple groups (since they are minimal normal), so they are uniquely determined by their composition factors.

(There might be a more elegant argument, but this is the first thing that came to mind.)

  • $\begingroup$ Hi, glad to meet you! $\ddot\smile$ why “$N_1\cong N_2$ does imply that $N_1$ and $N_2$ have the same multiset of composition factors”? $\endgroup$ – Benny Mar 16 '18 at 2:54
  • 1
    $\begingroup$ That's not what I said! What you say is even more obvious though. If $G$ is a finite group, and $N$ is a normal subgroup, then the multiset of composition factors of $G$ is the (multiset) union of the composition factors of $N$, and the composition factors of $G/N$. (This is fairly straightforward from the definitions.) This means that you can deduce the composition factors of $N$ from the composition factors of $G$ and $G/N$. So If $G/N_1$ and $G/N_2$ have the same composition factors (for example if they are isomorphic), then so do $N_1$ and $N_2$. $\endgroup$ – verret Mar 16 '18 at 3:11

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