If every right transversal to $H$ in $G$ is also a left transversal to $H$ in $G$, then $H$ is normal in $G$. Let $H\leq G$. A subset $T$ is said to be a right transversal to $H$ in $G$ if $T$ contains just one element from each right coset of $H$ in $G$. > If every right transversal to $H$ in $G$ is also a left transversal to $H$ in $G$, then $H$ is normal in $G$. I want to show that for all $g\in G, Hg=gH$. Let $T=\\{t_i\\}_{i\in I}$ be a transversal to $H$ in $G$. Then $\\{Ht_i\\}$ and $\\{t_iH\\}$ partitions $G$. But I can't proceed further as I can't show that $Ht_i=t_iH$. Note that the assumption given does not mean that every right coset of $H$ in $G$ is also a left coset of $H$ in $G$.

Suppose that $H$ is not normal in $G$. Then there exists $g \in G$ and $h \in H$ such that $k := g^{-1}hg \
ot\in H$.

So $g$ and $gk$ are in distinct left cosets of $H$ and hence can be extended to a left transversal of $H$ in $G$. But $gk=hg$, so $g$ and $gk$ are in the same right coset, and hence cannot be extended to a right transversal.

So if every left transversal is a right transversal then $H$ must be normal in $G$.