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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.

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Formulas spin out of control. Add backquotes around formulas with underscores or asterisks in them and it will work. –  Greg Kuperberg Nov 24 '09 at 3:38
    
@Greg, I don't see anything wrong here. –  Scott Morrison Nov 24 '09 at 4:00
    
Maybe it's browser-dependent? It works at home but not at school. –  Greg Kuperberg Nov 24 '09 at 6:48
    
In your first line, maybe you want E(g) = g if g\in H? –  Theo Johnson-Freyd Nov 24 '09 at 17:29
    
yes, thanks. i fixed it. –  Dave Penneys Nov 24 '09 at 18:40
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1 Answer

up vote 7 down vote accepted

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, §3, Chap. I, in the book The Theory of groups by H. Zassenhaus) That should do it, no?

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Fantastic! I will check out the book. Offhand, do you know if the result still holds if $G$ is infinite as $H$ is of finite index? –  Dave Penneys Nov 24 '09 at 5:58
    
You only need $H$ to be of finite index, so $G$ can be as infinite as you like. Interestingly, the result is also true if $H$ is finite. –  Mariano Suárez-Alvarez Nov 24 '09 at 6:18
    
The proof can be previewed on google books. It is not clear for me what a quadratic matrix is? Does it mean it satisfies a quadratic equation? Degree of A means the rank of A? –  Sebastian Burciu Jan 3 '10 at 13:38
    
@Sebastian: hmm? –  Mariano Suárez-Alvarez Jan 3 '10 at 17:14
    
It is essentially Hall's Marriage problem. –  Steve D Mar 23 '10 at 10:37
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