I am working on proving the following fact:
Let $G$ be a finite group and $H$ be a subgroup. Then there exists a complete set of left coset representatives for $G/ H$ which also form a complete set of right coset representatives for $H \backslash G$.
Here is the proof so far: Let $x\in G$. Then the double coset $HxH$ can be written as either a union of the left cosets $L=\{(hx)H: (h\in H)\}$ or as a union of the right cosets $R=\{H(xh):(h\in H)\}$. At this point I assume the following fact, which I'm struggling to prove:
For $L$ and $R$ as above, $|L|=|R|$.
Under the assumption, letting $r=|L|=|R|$, we can number the cosets in $L$ and $R$ as $L=\{(h_ix)H\}_{i=1}^r$ and $R=\{H(xh_i')\}_{i=1}^r$. Notice that $h_ixh_i'\in (h_ix)H\cap H(xh_i')$ for each $i$, so that the set $\{h_ixh_i'\}_{i=1}^r$ is a complete set of representatives for all the left cosets in $L$ and all the right cosets in $R$.
Since double cosets partition $G$, we can repeat the process for all double cosets to form a complete set of representatives for all the left cosets and right cosets of $H$ simultaneously.
Any hints on proving $|L|=|R|$? I was thinking of using the orbit-stabilizer theorem, but was having some trouble getting it right. It could also be something easy that I'm just not seeing.
Since distinct left cosets are disjoint and distinct right cosets are disjoint, and each coset has size $|H|$, then since $\bigcup L = HxH=\bigcup R$, we get $|L|= |HxH|/|H| = |R|$.