# Proof that this is a normal subgroup

Let $\phi: G$ to $J$ be a homomorphism onto all of $J$. Let $H$ be a normal subgroup of $G$ and let $K=\phi(H)$ be the image of $H$ under the homomorphism.

$1$. Show $K$ is a subgroup of $J$.

$2$. Now show $K$ is normal of $J$. Also find an example to show this can fail if $\phi$ is not onto.

$1$. Proof: We must show closure and inverses. So for $k_1,k_2 \in K$, $k_1k_2\in K$. So $k_1=\phi(h_1)$ and $k_2=\phi(h_2)$. Since $\phi$ is a homomorphism, $\phi(h_1)\phi(h_2)=\phi(h_1h_2)$. But $h_1h_2\in H$ since $H$ is a normal subgroup of $G$. Also $K=\phi(H)$ be the image of $H$ under the homomorphism, thus $k_1k_2 \in K$. Next we show inverses. So for $k_1\in K,$ show $k_1^{-1} \in K$. Since $k_1\in K,$ we know that $k_1=\phi(h_1)$, thus $h_1\in H$, so $h_1^{-1}\in H$. Therefore $\phi(h_1^{-1})$=$k_1^{-1} \in K$. So $K$ is a subgroup of $J$.

$2$. We must show that $K$ is normal of $J$. So for all $j \in J$, $jK=Kj$ or any other normal properties. Any idea on which one I might show? Also any idea on what an example would be?

Let $j\in J$, $k\in K$. Then, $j=\phi(g)$, for some $g\in G$, and $k=\phi(h)$ for some $h\in H$.
Then $jkj^{-1}=\phi(g)\phi(h)\phi(g)^{-1}=\phi(ghg^{-1})$. Since $H$ is normal $ghg^{-1}\in H$, so $jkj^{-1}\in\phi(H)=J$ q.e.d.
Hint: let $j \in J$, and let $g \in G$ such that $\phi(g) = j$ (note $g$ must exist since $\phi$ is onto). Let $h \in H$. Then $j^{-1}\phi(h)j = \phi(g)^{-1}\phi(h)\phi(g) = \phi(g^{-1}hg)$. Now what do you know about $gh^{-1}g$?