Properties of normal subgroup I am reading about normal subgroups and see that if $f:G\to G'$ is group homomorphism, and $H=\text{ker}(f)$, then for $x\in G$ we have
$$xH=Hx$$
I do not know how to read this notation.  I want to pull from it the property that they are both equal to $xH=Hx=f^{-1}(f(x))$ but I don't know where to begin.  This is the first time I have seen an element in composition with an entire set (beginner) so could you give me a full explanation?  Thanks.
 A: Typically (i.e. unless explicitely stated otherwise) in group theory if $A,B\subseteq G$ then
$$AB:=\{ab\ |\ a\in A, b\in B\}$$
with the special case (when one of the sets is a singleton)
$$xH:=\{x\}H$$
$$Hx:=H\{x\}$$
Note that since the multiplication in $G$ is associative then so is the multiplication of subsets.
Now if $H\subseteq G$ is a subgroup such that $xH=Hx$ for all $x\in G$ then we say that $H$ is normal. So what you are trying to prove is that the kernel of any homomorphism is a normal subgroup.
The property $xH=Hx$ means that you need to show two inclusions. In other words you have to show that for any $h\in H$ there exists $h'\in H$ such that $xh=h'x$ and vice versa.
Lets focus on "$\subseteq$" inclusion. As you can see $xh=h'x$ if and only if $xhx^{-1}=h'$. So the property can be rephrased as: for any $h\in H$ the element $xhx^{-1}$ belongs to $H$. But $H$ is the kernel of $f$ so
$$f(xhx^{-1})=f(x)f(h)f(x^{-1})=f(x)ef(x^{-1})=f(x)f(x^{-1})=f(xx^{-1})=f(e)=e$$
meaning $xhx^{-1}\in H$. Thus $xHx^{-1}\subseteq H$ or equivalently $xH\subseteq Hx$. The other inclusion is proved analogously.
