# Same left KG-module induced by nonisomorphic G-sets

For any field $K$, a finite group $G$, and left $G$-set $X$, $KX$ has a obvious left $KG-$ module structure, where $KG$ is the usual group algebra generated by the group $G$.

The question is

Find two non-isomorphic $G$-sets $X$ and $Y$ for which $KX\cong KY$. Especially, do this for $K=\mathbb{C}, G=\mathbb{Z}/{2\mathbb{Z}}$, and $X,Y$ are finite sets.

Restrict to the case $K=\mathbb{C}, G=\mathbb{Z}/{2\mathbb{Z}}$, and $X,Y$ are finite sets.

After some attempts, I think the points are the following:

1, $X\not\cong Y$ as $G$-sets iff they have different orbit structure, since $G$ has two elements, suppose it is generated by $\tau$ so every $G$-set has orbits of two types: orbit with one element $x$, so $\tau(x)=x$, and orbit with two elements $y, \tau(y)$.

2, $KX\cong KY$ may be possible only if $X, Y$ have the same size, and since we can map every element $x_i$ in $X$ to a linear combination of $\sum_{j=1}^mk_{i,j}y_i\in KY$, not necessarily only $y_i$. So we only need to find the possible choices of values on orbit space of $X$ to make sure the matrix the homomorphism determined is invertible.

I have used the above idea (I think they are correct) to construct examples, and I have rulled out the cases $|X|=|Y|=2, 3, 4$, but still got stuck in dealing with higher dimensional cases.

Can anyone give me some hint how to proceed further? or are there any points that I failed to see till now?

Thanks in advance!

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Hi, I've removed the "seek-hint" tag and replaced with the "homework" tag. (Whether there should be a hint tag was discussed before. I added a "homework" tag because people will often be more likely to leave a "hints only" answer like you asked for when the question is tagged as such.) –  Willie Wong Dec 14 '12 at 11:24
@Willie, thanks! –  ougao Dec 14 '12 at 13:24

## 1 Answer

If this really is a homework problem, then it's a rather mean one. Such two pairs do not exist for $K=\mathbb{C}$ and $G$ a cyclic group. Here are some general facts about such pairs of sets, which by the way are called Brauer relations:

• Any $G$-set is a disjoint union of transitive $G$-sets. The isomorphism classes of transitive $G$-sets are in bijection with conjugacy classes of subgroups of $G$. In one direction, you assign to a transitive $G$-set the point stabiliser of any point, and conversely, for $H\leq G$, take the $G$-set consisting of cosets $G/H$. So now, let's write $X=\sqcup_i G/H_i$, and $Y=\sqcup_j G/H_j'$. For such sets, we have $\mathbb{C}[X]=\bigoplus_i \mathbb{C}[G/H_i]$, and similarly for $Y$.

• The number of transitive $G$-sets is, as we have just seen, equal to the number of conjugacy classes of subgroups of $G$, so in a sense, the space of all $G$-sets has dimension equal to the number of conjugacy classes of subgroups (this can be made precise, i.e. there really is a space there, or rather a free $\mathbb{Z}$-module). On the other hand, one can show using basic character theory that the space of irreducible rational characters of $G$ has dimension equal to the number of conjugacy classes of cyclic subgroups of $G$. Note that any permutation representation is defined over $\mathbb{Q}$. Moreover, two permutation representations are isomorphic over $\mathbb{C}$ if and only if they are already isomorphic over $\mathbb{Q}$.

• From these two facts, one can show that the space of all essentially different pairs of $G$-sets $X,Y$ for which $\mathbb{Q}[X]\cong \mathbb{Q}[Y]$ (equivalently, $\mathbb{C}[X]\cong \mathbb{C}[Y]$) has dimension equal to the number of conjugacy classes of non-cyclic subgroups of $G$. The above computation shows that this space is at least this big. Equality is essentially the content of Artin's induction theorem: every rational representation is, up to a scalar multiple, a virtual permutation representation.

• When you are looking for such pairs the way you have done, there is another very helpful observation you should make. Every transitive subset $G/H$ of $X$ will contribute exactly one copy of the trivial representation in $\mathbb{C}[G/H]$ (why?). So to have any hope of getting $\mathbb{C}[X]=\mathbb{C}[Y]$, you at least want the trivial representation to cancel, i.e. you want $X$ and $Y$ to have the same number of transitive subsets. In the notation of the first bullet point, $i=j$.

• See here for a much more in depth discussion of Brauer relations.

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thank you so much!!!!!!! It is not a homework, the user Willie Wong suggests to lable this tag, because I only want to seek a hint on this problem. In fact, it is an old qualifying exam in our university, and I have never expected that it is so difficult.. –  ougao Dec 14 '12 at 19:28
You are very welcome. –  Alex B. Dec 14 '12 at 21:10