Minimal normal, maximal and isomorphic

Question

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.

Answer 1

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.)

Answer 2

The issue I find with this other answer is not that it isn't elegant but that it uses the concept of composition factor as well as the Jordan-Hölder Theorem, both of which come next in the book.

A more appropriate approach is to use that $N_1\subseteq N_2L$ to define $\phi\colon N_1\to N_2$ as $$ \phi(y_1)=y_2\iff y_1y_2\in L. $$ It is fairly easy to check that $\phi$ is a well-defined function (consider the relation ‘$\sim$’ in $N_1\times N_2$ by $y_1\sim y_2$ iff $y_1y_2\in L$, and then prove that it is indeed a function). Moreover, given that the elements of $N_1$ commute with the elements of $N_2$, one can verify that $\phi$ is in fact a homomorphism. Showing that $\phi$ is mono is straightforward. It is epi basically because $N_2\subseteq LN_1$.

Note finally that the hypotheses can be relaxed in several ways.