Let $G$ and $X$ be groups with a surjective homomorphism $\phi : G \to X $. Show that if $H \trianglelefteq G$ then $\phi(H) \trianglelefteq X$

Question

I proceeded as follows: To show $\forall \bar{g} \in G, \bar{g^{-1}} H \bar{g} \subset H $

Let $$z \in \bar{g^{-1}} H \bar{g} \\ z=\bar{g^{-1}}h\bar{g} \ , \exists h\in H \\ $$ Since $\phi$ is surjective, $ \forall \bar{g} \in \phi(G), \exists g \in G: \phi(g)=\bar{g}. $ Furthermore, $\bar{g^{-1}}=\phi(g^{-1})$

Hence it's enough to show $\phi(g)\phi(H)\phi(g^{-1})\subset \phi(H)$. Since $\phi$ is a homomorphism this simplifies to $\phi(gHg^{-1})\subset \phi( H)$

Also $H \trianglelefteq G$ thus $\forall g \in G , g^{-1} H g \subset H $

Letting $$z \in \phi(gHg^{-1}) \\ z=\phi(ghg^{-1}), \exists h \in H \\ \text{but } ghg^{-1} \in H\\ \to z \in \phi(H) \to \phi(gHg^{-1})\subset \phi(H)$$ Therefore $\phi(H) \trianglelefteq X$

Is the proof correct?

Answer 1

We have to prove that for every $x \in X$ it is $x^{-1}\phi(H)x \subseteq \phi(H)$. Since the homomorphism is onto, for every $x \in X$, it does exist $g \in G$ such that $x=\phi(g)$. But then, $x^{-1}\phi(H)x=\phi(g)^{-1}\phi(H)\phi(g)=\phi(g^{-1})\phi(H)\phi(g)=\phi(g^{-1}Hg)\subseteq \phi(H)$, because $g^{-1}Hg \subseteq H$ by hypothesis.