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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:

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:

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?

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See what the application $f(x)=ax$ does with the elements of the group. That is the permutation corresponding to $a$. Do this for every $a$ to find the permutation representartion. – Beni Bogosel Jul 13 '14 at 15:17
@BeniBogosel The OP seems to understand this; the question is why the MathWorld page has the statement "not all groups are representable as permutation groups." – angryavian Jul 13 '14 at 15:28
@angryavian: The phrase: "I couldn't figure out how to do it" says the opposite. The wolfram citation may be wrong, since it contradicts Cayley's theorem. – Beni Bogosel Jul 13 '14 at 15:32
The quoted statement from mathworld seems to be just wrong. – Andreas Blass Jul 13 '14 at 15:32
btw the quaternion group can't be represented faithfully as a permutation group on less than 8 letters. Among the 5 groups of order 8, it only shares this property with the cyclic group of order 8. – YCor Jul 16 '14 at 20:51
up vote 6 down vote accepted

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.

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Added some text to my answer. – Nicky Hekster Jul 13 '14 at 15:44

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.

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$Q_8$ 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 $\Omega$.

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