Can We Identify Quotient Groups as Subgroups of the Original Group?

[Hello Kitty]

Let G be a group and let N\trianglelefteq G, M\trianglelefteq G be such that N \le M. I would like to know if, in general, we can identify G/M with a subgroup of G/N.

Of course the obvious way to proceed is to look for a homomorphism from G to G/N whose kernel is M, but I can't think of one.

What I actually want to show is a more specialized result (namely the case when finite G/N is the nilpotent quotient of G and G/M is a maximal p-quotient of G for some p dividing the order of G/N) but the above is a lot cleaner and didn't yield obviously to a proof or counter-example so I thought I'd explore that first.

 

[fresh_42]

I would look for a counterexample, although I found none as the small groups are all "too cyclic". We have
$$
M\rtimes G/M \cong G \cong N \rtimes G/N\text{ and } N \triangleleft M
$$
This means we have to look for the automorphisms and need a group, where $G/M$ operates differently on $M$ than $G/N$ does on $N$. Therefore we need automorphism groups which have a few elements to select from.