Show that if N, K are normal subgroups of a group G, and N contains K then we have: $ G / N \cong (G/K) / (N /K) $

Question

Show that if $N, K$ are normal subgroups of a group $G$, and $N$ contains $K$ then we have: $$ G / N \cong (G/K) / (N /K) $$

Intuitively it looks correct, would like to know how I can approach this.

Answer 1

We can define a natural homomorphism from $G/K$ to $G/N$: $$ \varphi: G/K\to G/N,\quad gK\mapsto gN \quad\text{for}\ \ \ g\in G$$ First we verify that $\varphi$ is well-defined:

If $g_1^{-1}g_2\in K$, then clearly $g_1^{-1}g_2\in N$ since $ K\leqslant N $. So $\varphi$ is well-defined.

It should be clear that $\varphi$ is onto because $gK$ is the preimage of $gN$ for every $gN$ in $G/N$.

Finally, it suffices to show that the kernel of $\varphi$ is $N/K$. Since $\ker\varphi=\varphi^{-1}(N)$ which is exactly $N/K$. By the Fundamental theorem on homomorphisms we are through.