In another posting, there was a question about the following:
Let $G$ be a finite non-trivial group with the following two composition series:
$\{e\} = M_0 \triangleleft M_1 \triangleleft M_2 = G$
$\{e\} = N_0 \triangleleft N_1 \triangleleft \cdots \triangleleft N_r = G.$
Prove that $r=2$ and that $G/M_1 \cong G/N_1$ and $N_1/N_0 \cong M_1/M_0$.
The person posting the question went on to state that "By the second isomorphism theorem I know that $M_1N_2/N_2 \cong M_1/(N_2\cap M_1)$"
Here is my question. In order to use the second isomorphism theorem with $M_1$ and $N_2$ don't we need to know that $M_1 \leq N_G(N_2)$? And if so, then how do we know that $M_1$ actually is in the normalizer of $N_2$ in $G$?
If I can get over this hurdle, I understand the remainder of the original posting. Perhaps this is something obvious, but please help me see whatever it is that I am missing.
Thanks in advance.