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. 
 A: $\phi(g^{-1}xg)=(\phi(g))^{-1}e\phi(g)=(\phi(g))^{-1}\phi(g)=e$
A: If $x\in\ker\phi$, then $\phi(x)=1$, so $\phi(g^{-1}xg)=\phi(g^{-1})\phi(x)\phi(g)=\phi(g)^{-1}1\phi(g)=1$.
A: 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.
A: All the above proofs are correct, but maybe the most elegant way to say this is to resort to the first isomorphism theorem which says in particular that kernels of homomorphisms are normal subgroups.
