Relation between left and right coset representatives of a subgroup Let $G$ be a finite group and $H$ a subgroup. Is it true that
a set of right coset representatives of $H$ is also set of left coset 
representatives of $H$?
 A: No this not true unless $H$ is a normal subgroup. Suppose that $H$ is normal. Then for all $a \in G$, $aHa^{-1} = H$ so $aH = Ha$. Suppose that the set of left and right cosets are the same. Since cosets are partitions of $G$, $aH = Ha$. Hence $aHa^{-1} = H$. Since this holds for all $a \in G$, $H$ is a normal subgroup. 
However for any subgroup $H$ of $G$ it is true that the set of left cosets and right cosets are in bijection. Namely $\Phi(aH) = Ha^{-1}$. 

Let $R$ and $L$ be the set of right or right and left cosets. Above I showed that $R = L$ if and only $H$ is normal .
Now suppose that $H$ is not normal. So $R \neq L$. Hence there is a right coset $\sigma$ that is not a left coset. Since you assumed that the groups are finite, the size of each left and right cosets are equal. Hence if $\sigma$ is not any left coset, then $\sigma$ must intersect two left cosets $\tau_1$ and $\tau_2$. Let $a_1 \in \sigma \cap \tau_1$ and $a_2 \in \sigma \cap \tau_2$. Now form a set of representatives $K$ for left cosets, where $a_1$ and $a_2$ are choosen representative for $\tau_1$ and $\tau_2$ respectively. Now $K$ is not a set of representative for right cosets since $a_1$ and $a_2$ are representative for the same right coset. So some right coset must have no representative. 
A: Not every left transversal is also a right transversal. The Group Properties Wiki has a list of subgroup properties that are stronger than "having [at least one] left transversal that is also a right transversal"; among these is normality as William notes.
However the left coset representatives' multiplicative inverses form a right transversal, because
$$\begin{array}{c l}xH=yH & \iff y^{-1}xH=H \\ & \iff y^{-1}x\in H \\ & \iff (y^{-1}x)^{-1}=x^{-1}y\in H \\ & \iff Hx^{-1}y=H \\ & \iff Hx^{-1}=Hy^{-1}. \end{array}$$
