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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?


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'm suspicious, since it seems like if you had a group $G$ that was transitive and not $2$-transitive on $X$ then this would fail for $X \times X$. – Dylan Moreland Jun 4 '12 at 20:20
Is $W$ supposed to be acting on $K$ and $L$? Is it acting on $K\times L$? You can act on $K\times L$ without acting on $K$ and $L$ separately. For example, if $K=L$, $W=S_2$, then $W$ can act on $K\times L$ by permuting the coordinates, $\sigma(a,b) = (b,a)$ when $\sigma\neq\mathrm{id}$. – Arturo Magidin Jun 4 '12 at 20:22
@ArturoMagidin I agree that the question could use clarification. I assumed that $K$ and $L$ began life as $W$-sets and that the action on $K \times L$ was just $w(k, l) = (wk, wl)$. Maybe it should act on the right, given the notation. – Dylan Moreland Jun 4 '12 at 20:27
@AndreasPi: Perhaps you can quote what you "found in the literature" with enough context? I wouldn't even be sure what $K/W$ means if $K$ is a $W$-set, though I suspect it would probably mean the set of orbits of $W$ in $K$. – Arturo Magidin Jun 4 '12 at 20:29
I have edited the post. I hope it makes the problem clearer. – Hamurabi Jun 4 '12 at 20:48
up vote 3 down vote accepted

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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Thanks Geoff! But how can it be established that the first component is one of the orbit representatives on K and the second component is ANY element of L? And how would it lead in the second case to simply L/Stab(k)? What do I miss in the picture? – Hamurabi Jun 5 '12 at 6:30
Consider in the free action case how $(k,l)$ and $(k',l')$ can be in the same $W$-orbit. There must certainly be an element $w \in W$ such that $kw= k'.$ Since the action of $W$ on $K$ is free, this $w$ is unique. Since we are looking at the $W$-orbits on $K \times L,$ we may therefore reduce to the case that $k = k'.$ So when are $(k,l)$ and $(k,l')$ in the same $W$-orbit? We need $w \in W$ such that $kw = k$ and $lw = l'.$ But as the action of $W$ on $K$ is free, the condition on the first component forces $w = 1$ so now $l = l'.$ – Geoff Robinson Jun 5 '12 at 9:13
In other words (in the free action case), when calculating the orbit representatives for the action of $W$ on $K \times L,$ we are free to pick a representative for the orbit of the first component. But one we fix the representative in the first component, elements with distinct second components (and that fixed first component) are in different orbits. The analysis when the actions are not necessarily free are similar, using the observation given in my answer. – Geoff Robinson Jun 5 '12 at 11:27
ok. so in the non-free version one has kw=k giving you the stabilizer subgroup of K with respect to W, which gives you w's that will act like lw=l′, which gives you L/Stab_W(K)? Is that right? – Hamurabi Jun 5 '12 at 11:54
I think you mean $L/{\rm Stab}_{W}(k)$ actually. You have to do this for each orbit of $W$ on $K.$ It is a little bit loose (at best) to write it as $K/W \times {\rm Stab}_{W}(K)$ if the book did that. I think it's really more precise to say $\bigcup_{k \in K/W} L/{\rm Stab}_{W}(k).$ The stabilizers of the individual $k$'s in different $W$ can be quite different in general. – Geoff Robinson Jun 5 '12 at 15:12

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