What if $G/M=G/N$.....? The question actually came from the following problem about direct product and minimal normal subgroups:

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:
Let $N,M$ be two arbitrary groups that satisfy the condition. First, $LN(\text{or }LM)=G$ since $L$ is maximal, and $N$ is not contained in $L$. Then by one Isomorphism Theorem, we have $$G/N= LN/N\cong L/(L\cap N)=L;$$
and also
$$G/M= LM/M\cong L/(L\cap M)=L.$$
Thus we have $G/N\cong G/M$. $~~ \square$ 

My question:
It's clear that we cannot infer $N\cong M$ from $G/N\cong G/N$; it yet seems to tell us that $N$ and $M$ have the same multiset of composition factors. (But how? It's precisely where I got stuck.)
If that is true, since every minimal normal subgroup can be written uniquely as the product of some simple groups, it will yield directly that the composition factors (some simple groups) of $M$ and $N$ are all the same, and hence $M\cong N$. 

PS: I asked a similar question yesterday, and have accepted the answer. But I found myself stuck at this point now......and a more elegant solution to my original problem, if there is, would be greatly appreciated!
 A: Here is a proof of a slightly more general result, which does not involve composition series.
Suppose that the finite group $G$ has two minimal normal subgroups $M$ and $N$ such that $G/M \cong G/N$. Then $M \cong N$. (I wonder whether this is true for infinite groups - probably not.)
Proof by induction on $|G|$. It is vacuously true for $|G|=1$ because then $G$ has no minimal normal subgroups.
The result is clear if $M=N$ so suppose $M \ne N$. Then $M \cap N = 1$ by minimality, so $MN/N \cong M$ and $MN/M \cong N$.
Now the group $G/M$ has minimal normal subgroup $NM/M$ $(*)$ isomorphic to $N$ with $(G/M)/(NM/M) \cong G/MN$, and similarly $G/N$ has minimal normal subgroup $MN/N$ isomorphic to $M$ with $(G/N)/(MN/N) \cong G/MN$. Since $G/M \cong G/N$ we can apply induction to $G/M$ and deduce that $MN/M \cong MN/N$ and hence that $M \cong N$.
You may need to think about the claim $(*)$ that $NM/M$ is a minimal normal subgroup of $G/M$. This follows from $N$ being minimal normal in $G$.
A: Consider the series 
$$1 \leq M \leq G \mbox{ and } 1\leq N\leq G.$$
The subgroups $M$ and $N$ are normal, and $G$ is finite, hence the series can be refined to sub-normal series (i.e. a series in which each term is normal in just-next term and quotient is simple.)
So we do refinement as follows: 
Since $G/N\cong G/M$, we have a refinement for them as 
$$1\leq M \leq M_2 \leq \cdots M_r=G \mbox { and } 1\leq N \leq N_2\leq \cdots \leq N_r=G$$
where $M_2/M\cong N_2/N$, $M_3/M_2\cong N_3/N_2$ and so on.
What remains is to do refinement below $M$ and $N$; by Jordan-Holder theorem, the totality of simple-quteitnts in both refinements is same (up to isomorphism+ordering).
But since the simple factors coming from terms above $M$ and $N$ are same with multiplicity, it follows that the simple factors coming from terms below $M$ and $N$ should be same with multiplicity.
I hope this suffice to give you the idea. Let me know.
