# If $\phi\colon G\to H$ is a homomorphism, and $x$ is in the kernel, prove $gxg^{-1}$ is in the kernel for all $g\in G$.

Let $G$ and $H$ be groups and let $\phi\colon G \to H$ be a homomorphism. Show that if $x \in \ker(\phi)$ then $g^{-1}xg\in \ker(\phi)$ for all $g \in G$.

** End question **

I know the definition of homomorphism means the operation is preserved

and the definition of kernel is $\ker(\phi)=\{g \in G : \phi(g) = I_H\}$

But I honestly have no idea how to start.
Any help would be appreciated.

• $\phi$ is a homomorphism, so $\phi(gxg^{-1})=\phi(g)\phi(x)\phi(g^{-1})$ doesn't it? I think you can probably take it from here. – postmortes Dec 5 '13 at 14:58

Saying that something is in the kernel means that applying $\phi$ to it gives $1$ (the identity element in the codomain). Thus take $g\in G$ and $x\in\ker\phi$; then $$\phi(gxg^{-1})=\phi(g)\phi(x)\phi(g^{-1})$$ and this is the first step. Now we use the fact that $x\in\ker\phi$: \begin{align} \phi(gxg^{-1})&=\phi(g)\phi(x)\phi(g^{-1})\\ &=\phi(g)1\phi(g^{-1}) \end{align} The $1$ can be removed: \begin{align} \phi(gxg^{-1})&=\phi(g)\phi(x)\phi(g^{-1})\\ &=\phi(g)1\phi(g^{-1})\\ &=\phi(g)\phi(g^{-1}) \end{align} and we can apply once more the property of $\phi$: \begin{align} \phi(gxg^{-1})&=\phi(g)\phi(x)\phi(g^{-1})\\ &=\phi(g)1\phi(g^{-1})\\ &=\phi(g)\phi(g^{-1})\\ &=\phi(gg^{-1}) \end{align} OK, what's $gg^{-1}$? It's $1$, of course, and you know that $1\in\ker\phi$, don't you? Go on: \begin{align} \phi(gxg^{-1})&=\phi(g)\phi(x)\phi(g^{-1})\\ &=\phi(g)1\phi(g^{-1})\\ &=\phi(g)\phi(g^{-1})\\ &=\phi(gg^{-1})\\ &=\phi(1)\\ &=1 \end{align} Here it is.
$\phi(g^{-1}xg)=(\phi(g))^{-1}e\phi(g)=(\phi(g))^{-1}\phi(g)=e$
If $x\in\ker\phi$, then $\phi(x)=1$, so $\phi(g^{-1}xg)=\phi(g^{-1})\phi(x)\phi(g)=\phi(g)^{-1}1\phi(g)=1$.