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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 $g_1,g_2 \in N$?

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The statement you are trying to prove, as described in your first sentence, is not true. For example, every finite group injects into a symmetric group, and while the former may have lots and lots of normal subgroups, the latter has exactly three. – Mariano Suárez-Alvarez Dec 26 '11 at 5:06
Do you know that the homomorphism is surjective? Then this would be true. – Dylan Moreland Dec 26 '11 at 5:07
Your proof as it stands has some errors. So, you could go through the answer given by @Alex. And, I thought I'll just point out some "misteaks" in your proof. First of all, you should see that there doesn't exist any element in the intersection $G \cap G/N$, as they are distinct sets and could be atmost isomorphic. And how is $g_1,~g_2 \in N$ coming into the picture, as you seem to say that they are in $G$. – user21436 Dec 26 '11 at 6:31

I will assume that you assume $\mathcal L$ is surjective and that your definition of a normal subgroup $N\leq G$ is a subgroup such that $gN = Ng$ for all $g\in G$. In this case, you need to prove that $g'\mathcal L(N) = \mathcal L(N)g'$ for all $g'\in G'$ (I assume you already know that the image of a subgroup is a subgroup).

To do this, let $g'\in G'$. Using the fact that $\mathcal L$ is surjective, we have some $g\in G$ such that $\mathcal L(g) = g'$, hence $g'\mathcal L(N) = \mathcal L(g)\mathcal L(N) = \mathcal L(gN) = \mathcal L(Ng) = \mathcal L(N)\mathcal L(g) = \mathcal L(N)g'$, thus $\mathcal L(N)$ is normal.

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thank u so much – Junior II Dec 26 '11 at 5:34
Dear Junior, if Alex was right in assuming you were assuming the map is surjective, please edit the question body to reflect this. – Mariano Suárez-Alvarez Dec 26 '11 at 5:35

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