Homomorphism between a group and its factor group 
Theorem. Let $H$ be a normal subgroup of $G$. Then $\gamma:G\rightarrow G/H$ given by $\gamma(x)=xH$ is a homorphism with kernel $H$.

My question is in proving that $H$ is indeed the kernel of $\gamma$. It says:

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

I undertstand the statement for $xH=H$, but why does this prove that $H$ is the kernel? Why does this say that $\gamma[H]=e'$, where $e'$ is the identity in the factor group $G/H$?
 A: Proof that $H\subset\ker\gamma$: 
If $h\in H$ then $\gamma(h)=hH=H$ and the identity in $G/H$ is $eH=H$.
Proof that $\ker\gamma\subset H$:
If $\gamma(g)=H$, then $gH=H$. Hence, $g=ge\in H$.
A: Because $e'=H$.
Explanation: you chose to denote as $e'$ the identity element of the factor group $G/H$. But if you review the definition of the factor group (or quotient group) and the multiplication operation on it, the identity element of $G/H$ is $H=eH$, the coset of the identity $e\in G$. (Of course, here $H$ is viewed as a single element of the group $G/H$.)
A: 1) $xH = H$ implies that $x* e = h$ for some $h\in H$, where $e \in H$ is the identity element.
2) The kernel of $\gamma$ are those $g \in G$ so that $gH = H$ (since $\gamma(g) = gH$, and the identity of the factor group is the coset $H$), so they consist of the $g \in H$, by $(1)$.
A: $\gamma$ goes from $G$, so it cannot take $H$ as an argument. Let $e$ be the identity of $G$; $H$ is the identity of $G/H$, so the definition of the kernel:
$$\ker(\gamma) = \{x \in G | \gamma(x) = e_{G/H}\}$$
becomes
$$\ker(\gamma) = \{x \in G | \gamma(x) = H\}$$ and $\gamma(x) = xH$, so we have instead
 $$\ker(\gamma) = \{x \in G | xH = H\}$$
So $x \in \ker(\gamma)$ if and only if $xH=H$. Is that clear?
