Suppose $G$ is a group and $H$ is a subgroup of $G$. Let's call $T \subset G$ a left/right transversal of $H$ iff it is a system of representatives of left/right cosets of $H$ respectively. Let's call $T$ a double transversal of $H$ iff it is both left and right transversal of $H$ simultaneously. Does a double transversal exist for every subgroup $H$ of $G$?
In case when $H$ is finite, a double transversal for $H$ can be built the following way:
Suppose $g \in G$ is an arbitrary element. Then the double coset $HgH$ can be represented as union of the right and left cosets from the collections $R = \{Hgh' | h' \in H\}$ and $L = \{hgH | h \in H\}$ respectively. Note, that because $\bigcup R = HgH = \bigcup L$, we have $|L| = \frac{|HgH|}{|H|} = |R| = t$. Suppose, $R = \{Hgh_i'\}_{i=1}^t$ and $L = \{h_igH\}_{i=1}^t$ Then $h_igh_i' \in Hgh_i \cap h_igH$. That means, that $\{h_igh_i'\}_{i=1}^t$ is both a system of representatives from $R$ and a system of representatoives from $L$.
Since double cosets partition $G$, we can repeat the process for all double cosets to form the double transversal we were trying to build.
That case is easily generalised to the one when $H$ contains a finite index subgroup $N$, which is normal in $G$, by applying the previous result to quotients. However, I have no idea what to do in case when $H$ has no such subgroup.
Best Answer
I think the subgroup $\langle a \rangle$ of the Baumslag-Solitar group $${\rm BS}(1,2) = \langle a,b \mid b^{-1}ab = a^2 \rangle$$ is a counterexample.
In this group, the left coset $bH$ is the disjoint union of two right cosets $Hb \cup Hba$, with elements $ba^{2k}$ in $Hb$ and elements $ba^{2k+1}$ in $Hba$.
But a common left and right transversal would have to contain elements from $Hb$ and $Hba$ and so it would contain two elements from $bH$.