Let $G_1$ and $G_2$ be groups and suppose $\phi: G_1\mapsto G_2$ is a homomorphism. Then $\ker (\phi)\unlhd G_1$.
Need some feedback and help proving this. I am still new to proofs, but here's my attempt.
Proof.
We want to show $\ker (\phi)$ is normal, therefore, we must show that for any $h\in \ker (\phi)$ and $g\in G_1$, then $ghg^{-1}\in \ker (\phi)$. Since $h\in \ker (\phi)$, then $\phi(h)=1$. Thus,
\begin{align} \phi(ghg^{-1})&=\phi(g)\phi(h)\phi(g^{-1})\\ &=\phi(g)\cdot 1\cdot\phi(g^{-1})\\ &=\phi(g)\phi(g^{-1})\\ &=\phi(g\cdot g^{-1})\\ &=\phi(1)\\ &=1. \end{align}
Hence, $ghg^{-1}\in \ker (\phi)$, which implies $\ker (\phi)$ is a normal subgroup in $G_1$.
You're right.
An alternative method, following yours right until the end, goes like this:
\begin{align} \phi(ghg^{-1})&=\phi(g)\phi(h)\phi(g^{-1}) \\ &=\phi(g)\cdot 1\cdot \phi(g^{-1})\\ &=\phi(g)\phi(g)^{-1}\\ &=1. \end{align}
The conclusion is, of course, the same.