Understand part of a proof to show that if K is normal in G, then K is the kernel of a homomorphism

Question

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?

Answer 1

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, $H=K$.

Answer 2

Your understanding is correct. To avoid using contradiction, you can have positive arguments.

If $x\in G$ is such that $xK = K$, then $x=x\cdot e \in K$ where $e$ is the identity element.

Conversely for $k \in K$, $kK \subseteq K$ because $K$ is closed under $\cdot$ operation as a subgroup. And if $k^\prime \in K$ then $k^\prime = k\cdot(k^{-1} \cdot k^\prime) \in kK$. proving that $kK=K$.

Answer 3

Remember that any subgroup $\;K\;$ of a group $\;G\;$ defines an equivalence relation $\;R\;$ on $\;G\;$ defined by $\;xRy\iff y^{-1}x\in K\;$. You can prove this by yourself, it's easy.

Now, the set of equivalence classes, denoted sometimes by $\;K\backslash G\;$, is a group which operation is $\;[x]\cdot[y]:=[xy]\,,\,\,x,y\in G\;$, iff $\;K\;$ is a normal subgroup. We usually write every equivalence class in the form $\;xK\;$ and not $\;[x]\;$ , then the operation between equivalence clases is $\;xKyK:=xyK\;,\;\;x,y\in G\;$, and we get then that $\;xK=yK\iff y^{-1}x\in K\;$ , and also $\;xK=x\iff x\in K\;$. In this case we denote this group of equivalence clases as $\;G/K\;$ and call it " the quotient group of $\;G\;$ by $\;K\;$"