What does this group permute?

In a question, I'm given a group $G=Q_1 \times Q_2 \times \cdots \times Q_t$, where each $Q_i$ is a generalized quaternion group, and told it is a transitive permutation group. But what set does the group permute?

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Some set. Generally, there are many sets on which a group can act transitively; for example, $G$ acts transtively on itself (by left multiplication), and on the set of left cosets of any subgroup; it acts faithfully and transitively on itself, it acts faithfully and transitively on the set of left cosets of a subgroup if and only if the subgroup is core-free. So the answer is "many different sets, all you know is that you are considering a particular set on which the group acts faithfully and transitively." – Arturo Magidin Nov 26 '11 at 5:23
Are your above results for a general group or my group? Also, if $X$ is this set, then how do I show $G_x=\langle 1\rangle$ for some $x\in X$? As in, how do I use this unknown set? – Zeophlite Nov 26 '11 at 5:28
If you insist on only giving us piece-meal information about the problem you are trying to solve, then it will be impossible to give you an answer. There are usually many sets on which a group acts faithfully and transitively; they are in bijection with the core-free subgroups of $G$: given any $X$, and $x\in X$, the action is isomorphic to the action of $G$ on the left cosets of $G_x$. So perhaps you are expected to show that all transtive faithful actions of $G$ are regular by using this? I don't know, because I don't know what your homework problem states. – Arturo Magidin Nov 26 '11 at 5:33
Apologies for compartmentalizing my question, I was trying to understand specific parts. I have to show that $G$ is regular. I think I need to show that for any subgroup, $H$, of $G$, the core of $H$ in $G$ is trivial. – Zeophlite Nov 26 '11 at 5:44
No, you need to show that the core of any nontrivial subgroup is nontrivial. – Derek Holt Nov 26 '11 at 17:50

We need one relevant fact about generalized quaternion groups: every element of order dividing 2 in a generalized quaternion group Q is in the center of Q. This is sometimes stated as “a generalized quaternion group has a unique element of order 2.”

Proposition: In a direct product G of generalized quaternion 2-groups (or any 2-group in which elements of order 2 are central), every non-identity subgroup contains at least two elements of the center of G, and in particular is not core-free.

Proof: Suppose G is a direct product of quaternion groups and let H be a non-identity subgroup of G. By Cauchy's theorem H has some element h of order 2. However, the components of h much each have order dividing 2, and so must each come from the center of that component's generalized quaternion group. Hence h itself must be in the center of G. Of course, the identity is the second such element. Now the subgroup generated by h has order 2 and is central, so normal in G, but then it is contained in the core of H in G and H is not core-free.

Corollary: The only transitive action of a direct product of generalized quaternion groups is the regular action.

Proof: A transitive action is specified up to isomorphism by a conjugacy class of core-free subgroups (by the orbit-stabilizer theorem). The only core-free subgroup of such a group is the identity subgroup, which specifies the regular action.

This holds more generally for any direct product of p-groups satisfying the condition that every element of order p is central (which includes abelian groups). Such groups have been studied in several situations and have properties roughly dual to that of powerful p-groups, though with p = 2 it is sometimes more reasonable to assume all elements of order 4 are central, though not in this case.

Itô (1955) studied the groups in which every element of prime order was central and showed that if such a group had odd order, then it was nilpotent (and so just a direct product of p-groups in which every element of order p is central). More generally, Buckley (1970) studied finite groups in which every faithful transitive representation is regular (that is, such that every minimal subgroup is normal). He showed that such a group of odd order must be super-solvable (which is the natural change: if every chief factor is central of prime order, then the group is nilpotent, while if every chief factor is prime order, the group is supersolvable). van der Waall (1976) gave examples of non-supersolvable groups (of even order) in which every faithful transitive permutation representation was regular. Johnson (1971) studied minimal degrees where one does not require the action to be transitive, and then the generalized quaternion groups and cyclic 2-groups are the groups in which the smallest faithful permutation representations is the regular representation.

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In your corollary, you state "A transitive action is specified up to isomorphism by a conjugacy class of core-free subgroups (by the orbit-stabilizer theorem)". Could you please explain how? – Zeophlite Nov 27 '11 at 9:36
@Zeophlite: In this situation, all you need is the definition of regular: transitive with the identity as the stabilizer. – Jack Schmidt Nov 27 '11 at 17:33