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This question already has an answer here:

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.

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

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    $\begingroup$ 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 $E/L\to T$ ? $\endgroup$ – Max Apr 23 at 7:34
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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.

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  • $\begingroup$ Thank you very much. $\endgroup$ – Ilan Aizelman WS Apr 23 at 7:58
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    $\begingroup$ @IlanAizelmanWS You are welcome. $\endgroup$ – Bach Apr 23 at 8:06

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