# Non-character way to find the two dimensional representation for $Q_8$?

Let $Q_8=\langle \epsilon,i,j \mid \epsilon^2=e,j^2=i^2=\epsilon, \epsilon j=j \epsilon, \epsilon i=i \epsilon, ij=\epsilon ji\rangle$

I was able to find the unique(up to isomorphism) two dimensional representation for $Q_8$ in $\mathbb C^2$ using the character of the representation. This was not too difficult to cook up (essentially just using that $f(\epsilon)=-id$), but I was hoping for a different way to find this representation, so my question is this:

If all I knew was that there is a $2$-dimensional representation for $Q_8$ would there be an easy way to derive it?

Here is a weird idea that I had:

Let $$S:= \{f: Q_8 \to \mathbb C\}$$ We can make $S$ into a vector space $\mathbb C$-vector space with pointwise addition. We could then define the action of $Q_8$ by $g f(x)=f(g^{-1}(x))$ but somehow I still need to put some constraints on the functions involved, to get a $2$-dimensional vector space. Perhaps there is a nice way to proceed in this line of thinking, but all recommendations are welcome.

• $S$ is what is called the regular representation, and it's not a weird idea. In fact it contains copies of every irreducible representation, for any finite group. The constraints you need are determined by characters, but you will get $2$ copies of the representation you want this way without a bit more effort. Commented Mar 21, 2017 at 16:40
• hmm. Could I mod out by functions that identify $\epsilon$ with $-1$? Commented Mar 21, 2017 at 18:23

$Q_8$, almost by definition, has a $1$-dimensional quaternionic representation. This forgets down to a $2$-dimensional complex representation by restriction of scalars.
• By" forgets down" do you mean for example, sending $\epsilon \mapsto -1$? This seems that it would work, or by restriction of scalars do you mean consider the inclusion $\mathbb C \hookrightarrow Q_8$ and consider the induced representation in this way? Commented Mar 21, 2017 at 18:19
• @Andres: I mean consider any inclusion $\mathbb{C} \to \mathbb{H}$ and restrict the action of $\mathbb{H}$ to the action of $\mathbb{C}$. (Induction means going the other way.) Commented Mar 22, 2017 at 3:33
Another way to find a faithful representation of dimension $2$ would be to use the classification of finite subgroups of $GL_2(\mathbb{C})$. Then the elements of order $1,2,4$ of $Q_8$ should map to matrices of order $1,2,4$ in $GL_2(\mathbb{C})$, which generate finite subgroups of order $1,2,4$. By the way, there is a nice and elementary proof that $Q_8$ has no faithful representation into $GL_2(\mathbb{Q})$, see here.