Quaternion Group as Permutation Group

Question

I was recently, for the sake of it, trying to represent Q8, the group of quaternions, as a permutation group. I couldn't figure out how to do it.

So I googled to see if somebody else had put the permutation group on the web, and I came across this:

http://mathworld.wolfram.com/PermutationGroup.html

Not all groups are representable as permutation groups. For example, the quaternion group cannot be represented in terms of permutations.

This strikes me as a very odd statement, because I quickly checked this:

http://mathworld.wolfram.com/CayleysGroupTheorem.html

Every finite group of order n can be represented as a permutation group on n letters

So it seems Q8 should be representable a sa permutation group on 8 letters.

How can these two quotes be reconciled?

Answer 1

Follow Cayley's embedding: write down the elements of $Q_8=\{1,-1,i,-i,j,-j,k,-k\}$ as an ordered set, and left-multiply each element with successively with each element of this set - this yields a permutation, e.g. multiplication from the left with $i$, gives you that the ordered set $(1,-1,i,-i,j,-j,k,-k)$ goes to $(i,-i,-1,1,k,-k,-j,j)$, which corresponds to the permutation $(1324)(5768)$. Etc. Can you take it from here? So it can be done and the statement on the WolframMathWorld - Permutation Groups page must be wrong.

Answer 2

If $G$ is a finite group and $H$ a subgroup of $G$, we can consider the action of $G$ on the set of left cosets of $H$ by left-multiplication, i.e., $g\cdot (aH) = (ga)H$. This action is trivially transitive. The stabilizer of $aH$ is $aHa^{-1}$, so the kernel of the action (i.e., the set of $g$ fixing every $aH$) is the intersection of all the $aHa^{-1}$, which is the largest normal subgroup of $G$ contained in $H$ — in particular, the action is faithful iff $H$ contains no non-trivial normal subgroup of $G$. Conversely, given a faithful transitive action of $G$ on some (finite) set $X$, if we chose $o\in X$ and call $H$ the stabilizer of $o$, every element of $X$ can be written $a\cdot o$ for some $a\in G$ and it is easy to see that the map from the set of left cosets $G/H$ to $X$ taking $aH$ to $a\cdot o$ is an isomorphism of $G$-sets.

In other words, every faithful and transitive (left) action of $G$ on some (finite) set $X$ can be seen as the natural left action of $G$ on the left cosets of some subgroup $H$ which contains no non-trivial normal subgroup (and clearly, this subgroup is determined up to conjugacy by the action — it corresponds to the choice of $o$ whose stabilizer is taken).

Now we can always take $H=\{1\}$, corresponding to the left-regular action. The thing about the quaternion group (and which was probably meant by the MathWorld quote) is that, since all its subgroups are normal (so every non-trivial subgroup contains a non-trivial normal subgroup, namely itself), the left-regular action is the only faithful and transitive action it admits.

Answer 3

$Q_8$ can be represented as below: $Q_8=‎\langle ‎\alpha,\beta‎\vert ‎\alpha^4=‎\beta‎^4=1, ‎\alpha\beta\alpha=\beta, ‎\beta^2=\alpha^2‎‎\rangle‎$. Now let $‎\alpha=(1247)(3685)$, $‎\beta=(1348)(2576)$ and $|\Omega|=8$. Then $Q_8$ is a permutation group on $\Omega$.

Answer 4

Not all groups are representable as permutation groups. For example, the quaternion group cannot be represented in terms of permutations.

For the record, the page you linked to doesn't say this anymore. I'm guessing they recognized it was an error and removed it some time in the last three years.