Let $G$ be group, $N$ is normal in $G$ and $\mathcal{L}:G \to G^\prime$ be a surjective group homomorphism, then prove that the image $\mathcal{L}(N)$ of $N$ is a normal subgroup of $G’$.
I let $g_{1},g_{2} \in G$ such that, $g_1,g_2 \in G/N$. Since $N$ is normal $g_1 N=N g_1$ and $g_2 N= N g_2$. $g_1 N g_2 N = g_1 g_2 N = N$, $N g_1 N g_2 = N g_1 g_2 = N$, $\mathcal{L}( g_1 g_2 N) = g_1^\prime g_2^\prime \mathcal{L}(N)= \mathcal{L}(N) g_1^\prime g_2^\prime$. Hence this is my proof, I need some corrections, and is it necessary to consider the fact $g_1,g_2 \in N$?