Let $H\leq G$ be an inclusion of finite groups. Define a map $E\colon \mathbb{C}[G]\to \mathbb{C}[H]$ to be the $\mathbb{C}$-linear extension of
$$
E(g)=\begin{cases}
g &\text{if } g\in H\\\
0 &\text{else,}
\end{cases}
$$
i.e., $E$ is the projection onto $\mathbb{C}[H]$. A finite subset $B\subset \mathbb{C}[G]$ will be called a left basis for $G$ over $H$ if
$$
x=\sum\limits\_{b\in B} b E(b^\ast x)
$$
for all $x\in \mathbb{C}[G]$, where $\ast$ is the anti-linear extension of the map $g\mapsto g^{-1}$. For an example, take $B$ to be a set of left-coset representatives. Similarly, we can define a right basis to be a finite subset $B\subset \mathbb{C}[G]$ such that
$$
x=\sum\limits\_{b\in B} E(x b^\ast)b
$$
for all $x\in\mathbb{C}[G]$.


Note that there exist groups for which there is a basis which is both a left and right basis, but $H$ is not a normal subgroup of $G$. One can take the subgroup of the symmetric group $S\_n$ ($n\geq 3$) which fixes $1$. Then a set of left and right coset representatives is given by 
$$
\{ (1 j)|j=1,\dots,n\}.
$$
Does there always exist a basis which is both a left and right basis, or are there inclusions of groups for which there is no simultaneous left and right basis?


The motivation for this question is another question from subfactor theory: if $N\subset M$ is a finite index, extremal $II\_1$-subfactor, does there always exist a Pimsner-Popa basis which is both a left and right basis? The subgroup subfactor is an example of such a subfactor, and the question posed above is a watered-down version of the subfactor question, where perhaps an answer is already known or more easily obtainable.



Do subgroups have "two sided bases"?
If $H$ is a subgroup of finite index in a group $G$, there is a subset $\mathcal B$ of $G$ which serves both as a set of representatives for the left cosets of $H$ in $G$ and as a set of representatives for the right cosets of $H$ in $G$. (See, for example, Theorem 3, §4, Chap. I, in the book *The Theory of groups* by H. Zassenhaus) That should do it, no?


