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

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.

## marked as duplicate by Chinnapparaj R, Shaun, Community♦Apr 23 at 7:59

• Suppose you have a group $E$ and a normal subgroup $L$, and another group $T$ do you know what you need to specify to define a morphism $E/L\to T$ ? – Max Apr 23 at 7:34

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.