Theorem 14.9

Statement:

Let $H$ be a normal subgroup of $G$. Then $\gamma :G\rightarrow G/H$ given by $\gamma (x)=xH$ is a homomorphism with kernel $H$.

Proof:

Let $x,y\in G$. Then,

(1)
\begin{align} \gamma (xy) = (xy)H = (xH)(yH) = \gamma (x)\gamma (y) \end{align}

so $\gamma$ is a homomorphism. Since $xH=H$ if and only if $x\in H$, we can see that the kernel of $\gamma$ is indeed in $H$.