A set the quaternion group could act on

[R.P.F.]

Homework Statement

Find the smallest integer n such that the quaternion group G has a faithful operation on a set Sof order n.

Homework Equations

The Attempt at a Solution

So the homomorphism between G and permutations of S is injective. which means the order of S_n is bigger than or equal to the order of G. so we should start from n=4. Is it possible to find a subgroup of S_4 that's isomorphic to G. I think the answer is no. so we go to n=5,6 and so on. I'm having trouble finding such a subgroup.

Thanks!

 

[micromass]

It is easy to find a faithful action from the quaternion group on a set with 8 elements. But you can't find an faithful action on a set with less then 8 elements.

To prove this, assume that G acts faitfully on X, and |X|<8.
From the orbit-stabilizer theorem follows for every x\in X that [G:Stab(x)]<8. Thus Stab(x) cannot be trivial.
This implies that \{-1,1\}\subseteq Stab(x) for every x. Thus \{-1,1\}\subseteq \bigcap Stab(x)=Ker
Thus the kernel of our action is nontrivial...

 

[jbunniii]

A subgroup of either S4 or S5 with order 8 is a 2-Sylow subgroup, and all such subgroups are isomorphic (by conjugation). Both S4 and S5 contain isomorphic copies of D8, the dihedral group of order 8, so neither can contain a copy of the quaternion group. So S6 is the smallest candidate.

 

[R.P.F.]

It is easy to find a faithful action from the quaternion group on a set with 8 elements. But you can't find an faithful action on a set with less then 8 elements.

To prove this, assume that G acts faitfully on X, and |X|<8.
From the orbit-stabilizer theorem follows for every x\in X that [G:Stab(x)]<8. Thus Stab(x) cannot be trivial.
This implies that \{-1,1\}\subseteq Stab(x) for every x. Thus \{-1,1\}\subseteq \bigcap Stab(x)=Ker
Thus the kernel of our action is nontrivial...

Oh it wasnt that easy to find a subgroup of S8 that's isomorphic to the quaternion group...but I figured it out. The proof you provided is very neat. Thank you so much!

 

[R.P.F.]

A subgroup of either S4 or S5 with order 8 is a 2-Sylow subgroup, and all such subgroups are isomorphic (by conjugation). Both S4 and S5 contain isomorphic copies of D8, the dihedral group of order 8, so neither can contain a copy of the quaternion group. So S6 is the smallest candidate.

We haven't got to sylow thms yet. But I will look it up. Thanks a lot! :)