Prove two subgroups have the same order given an equivalence relation. Let $X$ be the set of all subgroups of a group $G$. The following is an equivalence relation $H$~$K$ where $H=xKx^{-1}$.
Prove: If $H$~$K$, then $|H|=|K|$.  
I'm guessing $H$,$K$ are subgroups of a group $G$.  Other than that, I'm not quite sure where to start on this one.  I know that if $|H|=|K|$ then they have the same amount of elements in them but I'm not sure how to prove that given the equivalence relation.

Check the map $f : K \rightarrow H$ such that $f(a) = xax^{-1}$. It's obviously surjective homomorphism. So if it is injective we are done. Suppose that $a \in \operatorname{Ker}(f)$, then $xax^{-1} = e \Rightarrow xa=x $ (right multiplication by $x$) $\Rightarrow a = e$ (left multiplication by $x^{-1}$). Therefore $f$ is indeed injective, because $\operatorname{Ker}(f) = \{e\}$. Here is the desired bijection.
