Let $K$ be a normal subgroup of $G$. Then we can show that $K$ is the kernel of a homomorphism with domain group $G$.
I found a proof for this, but I don't fully get it yet. The proof works like this:
- Define a mapping $\phi$, such as:
\begin{align*}
\phi: (G, \circ) &\rightarrow (G/K, \cdot) \\
x &\rightarrow xK
\end{align*}
Where $G$ is the domain group and $G/K$ is the codomain group with set composition as binary operation. We know $G/K$ exists, since $K$ is normal in $G$.
- Then we need to show that $\phi$ is indeed a homomorphism. That part is clear, so I skip it.
- Assuming $\phi$ is a homomorphism, we now need to show that $Ker \phi = K$, as claimed.
Now in the book this is done like so:
\begin{align*}
Ker \phi &= \{ x \in G : \phi(x) = K \} \\
&= \{ x \in G : xK = K \} \\
&= K
\end{align*}
I don't fully grok why $\{ x \in G : xK = K \} = K$.
I think this is because of the closure property of the subgroup $K$ of $G$?
Because if there would be a $x \in G$ such that $xK \ne K$, then $x \not \in K$. Otherwise, if $x \in K$, but for some $k \in K$ we find $xk \not \in K$, then $K$ can't be closed under its binary operation.
Is that understanding correct?
Best Answer
Call $H=\{x\in G\mid xK=K\}$. If $x\in K$, then $xK=K$ and hence $x\in H$, whence $K\subseteq H$. Vice versa, if $x\in H$ then $xK=K$ and, in particular, $xK\subseteq K$; by definition, this means that $\forall k\in K, \exists k'\in K$ such that $xk=k'$; take $k=e$ to conclude that $x\in K$, whence $H\subseteq K$. By the double inclusion, $H=K$.