abstract algebra – Second Isomorphism Theorem and Jordan-Holder

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 $M_1N_2/N_2 cong M_1/(N_2cap M_1)$

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.