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](http://en.wikipedia.org/wiki/Quaternionic_representation)" 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](https://mathoverflow.net/questions/47492/which-groups-have-only-real-and-quaternionic-irreducible-representations?rq=1) is similar but in that post $G$ is infinite and there is not the emphasis on "all representations are degree 1".) 



Which finite nonabelian groups have all their quaternionic representations of degree one?
**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](http://en.wikipedia.org/wiki/ADE_classification)). 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.


