Question:
Assume that$$G=G_0\supset ...\supset G_{i}\supset G_{i+1}\supset...\supset G_n=\{e\}$$ is a composition series for G,and $N$ is a proper normal subgroup of $G$.
Prove that $$\frac{G_{i}\cap N}{G_{i+1}\cap N}\triangleleft\frac{G_i}{G_{i+1}}.$$
(in the sense that $\frac{G_{i}\cap N}{G_{i+1}\cap N}$ is the image of the injective homomorphism $\frac{G_{i}\cap N}{G_{i+1}\cap N}\to \frac{G_i}{G_{i+1}}).$
Attempt:
By applying the third isomorphism theorem we have $$\frac{G_i\cap N}{G_{i+1}\cap N}\triangleleft \frac{G_i}{G_{i+1}\cap N}$$
since $G_{i}\cap N\triangleleft G_i$.
But I have trouble showing that $\frac{G_i}{G_{i+1}\cap N}$ is a subgroup of $\frac{G_i}{G_{i+1}}$,since the inclusion failed to be a homomorphism(even not well-defined).
Note that $G_i\cap N\triangleleft G_i$, since $N\triangleleft G$. Since $G_{i+1}\triangleleft G_i$ as well, we have $(G_i\cap N)G_{i+1}\triangleleft G_i$.
By the correspondence theorems we know that $$\frac{(G_i\cap N)G_{i+1}}{G_{i+1}}\triangleleft \frac{G_i}{G_{i+1}}.$$
The Second Isomorphism Theorem tells us that if $M\triangleleft G$ and $K$ is any subgroup, then $\frac{K}{K\cap M}\cong \frac{KM}{M}$, via the map $k(K\cap M)\mapsto kM$.
Taking $M=G_{i+1}$ and $K=G_i\cap N$, we have $$\frac{(G_i\cap N)G_{i+1}}{G_{i+1}} \cong \frac{G_i\cap N}{(G_i\cap N)\cap G_{i+1}}.$$ But $G_i\cap N\cap G_{i+1} = N\cap G_{i+1}$ (since $G_{i+1}\leq G_i$). Therefore, we have $$\frac{G_i\cap N}{G_{i+1}\cap N} \cong \frac{(G_i\cap N)G_{i+1}}{G_{i+1}}\triangleleft \frac{G_i}{G_{i+1}},$$ where the isomorphism is the natural one, sending $g(G_{i+1}\cap N)$ to $gG_{i+1}$.