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 $G$.
 A: Suppose that $H$ is not normal in $G$. Then there exists $g \in G$ and $h \in H$ such that $k := g^{-1}hg \not\in H$.
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 $G$.
A: Fix arbitrary $a\in H$ and $k\in G$. Consider the right transversal 
$$T = \{t_g\} \text{ where } t_g = 
\begin{cases}
g & g\ne k \\
ak & g = k
\end{cases}$$
and note that since $T$ is also a left transversal, we must have $ak\in kH$.  Thus $k^{-1}ak \in H$ for any $k$, so $g^{-1}Hg\subseteq H$ for any $g\in G$. This implies that $H$ is normal.
A: Pick any $g\in G$. Consider a right transversal $\{g,t_i\}$ containing $g$ for some $t_i$'s. It is clear that $g\in Hg$.
Now $\{g,t_i\}$ is also a left transversal. Thus $g$ is in some left coset of $H$, $gH$ to be exact.
For any $w\in Hg$, note that $\{w,t_i\}$ is a right transversal, thus a left transversal too. Since $t_i$'s each unchanged go to respective same left coset as before, $w\in gH$.
