Proof of Jordan-Holder theorem

Question

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.

Answer 1

Let $p_2:G\rightarrow G/M_2$ be the projection, $p_2(N_r)$ is normal since $N_r$ is normal, we deduce that $p_2(N_r)=1, N_r\subset M_2$, since $G/N_r$ is simple, $q_r(M_2)=1$ where $q_r:G\rightarrow G/N_r$ is the natural projection, $M_2\subset N_r$, henceforth, $M_2=N_r$.

Apply the same argument to show that $N_{r-1}=M_1$ and $M_0=N_{r-2}$, and deduce $r=2$ since $M_0$ is simple.