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A finite group $G$ has a finite set of irreducible representations over the complex numbers. All of these representations are linear (that is, are maps in 1x1 complex matrices) if and only if $G$ is abelian. Moreover, if the group $G$ is not abelian, those representations which are linear can be described by replacing $G$ by $G/G'$ (where $G'$ is the commutator subgroup.)

I need a similar classification in which the field of complex numbers is replaced by the quaternion division ring. Is there a similar theory? Can I classify those $G$ for which the quaternionic representations are linear? (Is it possible to even identify a normal subgroup $K$ of $G$ for which I am guaranteed that $G/K$ is "quaternionic linear"?)

I've identified some articles on "quaternionic" or "symplectic" representations but most of these are concerned with infinite groups and assume quite a lot of theory that does not seem relevant to finite groups.

(This mathoverflow question is similar but in that post $G$ is infinite and there is not the emphasis on "all representations are degree 1".)

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You can use the classification of finite subgroups of $\text{Sp}(1) \cong \text{SU}(2)$, right? The normal subgroup $K$ should be the intersection of the kernels of all homomorphisms to these finite subgroups. That's not very explicit, unfortunately. –  Qiaochu Yuan Jun 6 at 20:21
@QiaochuYuan That normal subgroup is too small. For example, for subgroups of $SU(2)$, only cyclic and quaternion groups satisfy this condition, not the dihedrals or symmetries of the platonic solids. –  Ben Webster Jun 6 at 20:54

2 Answers 2

up vote 7 down vote accepted

EDIT: My previous answer was completely wrong. Here's a corrected version.

First, note that this property descends to subgroups: every irrep of a subgroup is a summand of a restriction from the bigger group. Thus if all the irreps of the bigger group are 1-d, all the irreps of the small group will be their restrictions with the same one coming up multiple times.

If $G$ has $n$ 1-dimensional quaternionic representations whose sum is faithful, then $G$ must embed into the product of $n$ number of copies of the unit quaterions (that is, $SU(2)$). Consider the image of $G$ in any one of these representations, and then consider the image in $SO(3)$; since it is a finite subgroup of $SO(3)$, it is either cyclic, dihedral (including $C_2\times C_2$), or $A_4,S_4$ or $A_5$ (see Wikipedia). However, if it's nonabelian, this group has a real irrep (an irrep over the real numbers whose endomorphisms are $\mathbb{R}$, which is thus absolutely irreducible) which isn't 1-d, so it tensoring gives an irrep over the quaternions that isn't 1-d, so it must be cyclic or $C_2\times C_2$. This shows that the original subgroup of $SU(2)$ is either abelian, or the classic quaternion group.

Thus, $G$ is a subgroup of a product of cyclic and quaternion groups, and as we argued above, all such subgroups have this property.

EDIT: It's also worth mentioning that this shows that the notion of a maximal quaternionic linear subgroup exists: it's the quotient by the intersection of the kernel of all maps to abelian or $Q_8$ groups.

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Maybe I have misunderstood the question, but isn't every representation of the quaternion group of order 8 linear over the quaternions? –  Mark Wildon Jun 6 at 21:48
Yes, I would have thought that every representation of the quaternion group of order 8 would be "linear" over the quaternions. If G = $\langle x, y : x^4=y^4=1, yxy^{-1}=x^3, x^2=y^2 \rangle$ then I have four reps over the rationals. In addition I guess I could map $x$ to $\pm i$, $y$ to $\pm j$ but are there more than four of these options possible? Do I lose control of my set of irreducibles? (I am so comfortable with the complex field that I have no good intuition with the strange noncommutative quaternion division ring!) –  Ken W. Smith Jun 7 at 0:59
To any future readers: the comments above refer to my (wrong) previous answer. –  Ben Webster Jun 7 at 13:30

The original question has been answered. As background information, I point out that S. Amitsur classified finite groups which can occur as subgroups of multiplicative groups of divison algebras. These include finite groups which have a faithful "one-dimensional" representation over the quaternions. As noted in Ben's answer, any such finite group (in the general division ring case) has cyclic Sylow $p$-subgroups for each odd prime $p$ and cyclic or generalized quaternion Sylow $2$-subgroups. It also satisfies other group-theoretic properties, such as elements of order $p$ and $q$ commuting when $p$ and $q$ are distinct primes.

In fact, finite groups which occur as subgroups of multiplicative groups of division algebras are Frobenius complements. We note that any generalized quaternion group already occurs - this is not in conflict with Ben's answer: if that group has order greater than $8,$ then it has a non-Abelian dihedral group as a homomorphic image, so not all its irreducible representations over the quaternions are $1$-dimensional. The only non-solvable group which occurs is ${\rm SL}(2,5),$ which does already occur inside the multiplicative group of the real quaternions ( as does each generalized quaternion $2$-group)- again, the fact that ${\rm SL}(2,5)$ has $A_{5}$ as a homomorphic image means that not all its irreducible representations over the quaternions are $1$-dimensional.

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