# Relationship Between Normal Subgroup/Ideal and the Kernel of a Homomorphism

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.

• Oops sorry you're right, typo. Fixed it.
– Max
Commented Dec 11, 2016 at 4:33
• It's a bit lengthy, but if you have some spare time then you might find it helpful to take a look through this blog post by Tim Gowers. Commented Dec 11, 2016 at 4:38

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$(\phi) = N$.

• Is it an "iff" situation? Or is it only that Ker($\phi$) implies normal subgroup?
– Max
Commented Dec 11, 2016 at 4:36
• @Max: it's indeed an iff situation. Consider the canonical quotient homomorphism $G \to G/N$. What is its kernel? Commented Dec 11, 2016 at 4:37
• Yes, it is iff as @AlexWertheim mentioned. I'll update the answer for completeness. Commented Dec 11, 2016 at 4:39
• I'm not quite seeing why the kernel of $f:$ $G \rightarrow G$/$N$ is $N$. Because the kernel of $f$ would be all the elements of $G$ mapped to $0$, right? How does that relate to $N$?
– Max
Commented Dec 11, 2016 at 4:48
• $0$ is the identity element of $G/N$. But $G/N$ is the quotient group. Its elements are cosets and its $0$ (identity) element is $N$. Commented Dec 11, 2016 at 4:52

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 $N$.