I am trying to prove 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 $N_1/N_0 \cong M_1/M_0$.
I know that if $r < 2$ we have a contradiction since $G$ is non-trivial and not simple (since $M_1 \triangleleft G$). However, I am having a hard time writing down a contradiction for when $r > 2$. We know that $N_{r-1} \triangleleft G$ and $G/N_{r-1}$ is simple, but how can I use the fact that $M_1$ and $G/M_1$ are simple? By the second isomorphism theorem I know that $M_1N_2/N_2 \cong M_1/(N_2\cap M_1)$, which means $N_2\cap M_1 = \{e\} $ or $N_2\cap M_1 = M_1$, since $M_1$ is simple. Can we also conclude, by the second isomorphism theorem, that $$N_2/(N_2\cap M_1) \cong N_2M_1/M_1 \cong G/M_1.$$ So if $N_2\cap M_1 = \{e\}$ then $ G/M_1 \cong N_2/(N_2\cap M_1) \cong N_2$ is simple, which is a contradiction since $\{e\} \ne N_1 \triangleleft N_2$? The part I am confused about is showing $N_2M_1/M_1 \cong G/M_1$. It feels like it should be true since $N_2$ and $M_1$ are maximal, but I don't have a good way of writing it down.