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My question is the following:

Suppose one has two sets $K$, $L$ and the group $W$. What is $(K \times L)/W$? Is it isomorphic to $K/W \times L/W$?

I have found something different in the literature and now I am lost.

Can anybody help me here please?

Thanks

edit: Thanks for the replies. Here it gets more precise. I have read the following: $W$ is supposed to act freely on $K$ and $L$. Its not explicitly given how it acts on $K \times L$, just that it does. Then it is given that $(K\times L)/W$ is isomorphic to $K/W \times L$ (!). In case W does not act freely on $K$, one is supposed to get $K/W \times L/(\text{stabilizergroup}(K))$.

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    I have edited the post. I hope it makes the problem clearer.2012-06-04

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Re the new version: if $W$ acts freely on $K$ and $L,$ then it does act freely on $K \times L$, for certainly no non-identity element of $W$ can fix any element of $K \times L$ under the standard action on the direct product. Then the only reasonable interpretation of the claim in the book is that the orbits of $W$ on $K \times L$ have representatives given by ordered pairs where the first component is one of the orbit representatives on $K$ and the second component is any element of $L.$ This is correct. Similarly, to work out representatives for the orbits of $W$ on $K \times L$ if neither action is free, it is just a question of working out the stabilizer in $W$ of a general ordered pair $(k,l)$, and this is clearly ${\rm Stab}(k) \cap {\rm Stab}(l).$

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    right. that makes fully sense to me. thank you for the detailed explanation!2012-06-05