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

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?

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 $$kK=K$$.

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, $$H=K$$.

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 $$\;K\;$$"