# Representation of $Q_8$ over $\mathbb{R}$

I'm trying to solve the following problem,

Give an example of a finite group $G$ and its irreducible representation $L$ over $\mathbb{R}$ such that the division algebra $Hom_G(L, L)$ is isomorphic to the algebra $\mathbb{H}$ of quaternions.

$Hom_G$ definition

I looked at the dimension $4$ irreducible representation of $Q_8$ over $\mathbb{R}$, which is the homomorphism from $Q_8$ to $GL_4(\mathbb{R})$ that maps elements of $Q_8$ to corresponding left multiplication. The matrix representation can be found here. (On that page search Four-dimensional irreducible representation over a non-splitting field)

I found this embedding of $\mathbb{H}$ into $M_4(\mathbb{R})$ and I've checked that these matrices commute with those matrix representation of elements of $Q_8$ (therefore in $Hom_G(L,L)$) but I haven't checked these are all of them and I'm a little lost here.

Could anyone please explain a better way of understanding/approaching this question than lucky guess+google search?

Giving a homomorphism $\phi \colon \mathbb{H} \to \text{Hom}_{Q_8}(L,L)$ of $\mathbb{R}$-algebras is equivalent to giving a map

$$\hat{\phi} \colon \mathbb{H} \times L \to L$$

which is $\mathbb{R}$-bilinear and additionally satisfies:

• $\hat{\phi}(hr,l) = \hat{\phi}(h,rl)$ if $r$ is a real number.
• $\hat{\phi}(h,ql) = q \hat{\phi}(h,l)$ if $q \in Q_8$.
• $\hat{\phi}(h_1h_2,l)= \hat{\phi}(h_1,\hat{\phi}(h_2,l))$ .

The correspondence is given by $\hat{\phi}(a,l) = \phi(a)(l)$.

It is easy to see that the multiplication of quaternions $\mathbb{H} \times \mathbb{H} \to \mathbb{H}$ satisfies these properties, that is $L = \mathbb{H}$, if we let $Q_8$ act on $L = \mathbb{H}$ by seeing $Q_8$ as the group of units of $\mathbb{H}$ and act on $L$ by multiplication on the left.

So $L$ should be $\mathbb{H}$, which as a vector space over $\mathbb{R}$ is just $\mathbb{R}^4$. When you translate the action of $Q_8$ on $\mathbb{H}$ to the corresponding action on $\mathbb{R}^4$, you get precisely the irreducible $4$-dimensional representation you mentioned.

When you translate the multiplication of quaternions $\mathbb{H} \times \mathbb{H} \to \mathbb{H}$ into a map $\mathbb{H} \times \mathbb{R}^4 \to \mathbb{R}^4$ it gives you a map $\mathbb{H} \to M_4(\mathbb{R})$. This map is the same you mentioned as well, and being an embedding gives you that the corresponding $\phi$ is injective. But checking that this is an embedding is easy if you think of it as representing the multiplication of quaternions.

Finally to see that it is surjective, if you have a map $f \colon \mathbb{H} \to \mathbb{H}$ which is $\mathbb{R}$-linear and the elements $1$, $i$, $j$ and $k$ act by multiplication on the left, then the map is left multiplication by $f(1)$.