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What exactly is the relationship between a normal subgroup/ideal and the kernel of a homomorphism? I understand that if $N$ (or $I$) is the kernel of some homomorphism of $G$, then $N$ is a normal subgroup of $G$. What I don't understand is why.
How does $gng^{-1}$ $\in$ $N$ for all $n \in N$, $g \in G$ relate to it being the kernel of a homomorphism? Thank you.
If $\phi:G\to G$ is a homomorphism, $n \in$ ker $(\phi)$, and $g \in G$ is arbitrary, then $\phi(gng^{-1}) = \phi(g)\phi(n)\phi(g)^{-1}= \phi(g)\phi(g)^{-1} = e,$ where $e$ is the identity element. Therefore the ker$(\phi)$ is a normal subgroup.
On the other hand, if $N$ is a normal subgroup of $G$, then the map $\phi: G\to G/N$ is a homormorphism with ker$(\phi) = N$.
Let $G$ be a group. A normal subgroup of $G$ is the same thing as the kernel of a group homomorphism whose domain is $G$.
If $K$ is the kernel of a homomorphism of groups $\phi: G \rightarrow H$, then $K$ is a subgroup of $G$. Suppose $g \in G$ and $k \in K$. Then $$\phi(gkg^{-1})= \phi(g)\phi(k) \phi(g^{-1})= \phi(g)1_H \phi(g)^{-1} = 1_H$$
and so $gkg^{-1} \in K$. Thus $K$ is a normal subgroup of $G$.
On the flip side, if $N$ is a normal subgroup of $G$, the set $G/N$ of left cosets of $N$ in $G$ is a group, and $g \mapsto gN$ is a group homomorphism $G \rightarrow G/N$ whose kernel is $N$.
