Homomorphism from $Q_8$ to $GL_2(\mathbb{C})$

Question

From Dummit and Foote Abstract Algebra, 1.6.26.

Define the map $\varphi$ from $Q_8$ to $GL_2(\mathbb{C})$ defined on the generators by $$ \varphi(i) = \begin{bmatrix} \sqrt{-1} & 0 \\ 0 & -\sqrt{-1} \\ \end{bmatrix} \quad \text{and} \quad \varphi(j) = \begin{bmatrix} 0 & -1 \\ 1 &0 \\ \end{bmatrix} .$$ Prove that this map extends to a homomorphism.

Does this mean prove $\varphi$ is a homomorphism on $Q_8$?

If so, then I can show

$$\quad \varphi(ij) = \varphi(i)\varphi(j), \quad \varphi(ii) = \varphi(i)\varphi(i), \quad \text{and} \quad \varphi(jj) = \varphi(j)\varphi(j),$$ which shows $\varphi$ is a homomorphism on the generators. I think that automatically makes it a homomorphism on $Q_8$, since $i,j$ generate $Q_8$.

Is that correct?

Answer 1

Yes . . . Almost.

You need to show $\phi(ji)=\phi(j)\phi(i)$, since $ij=k\neq -k=ji$.

But you could say that $-k$ is a scalar multiple of $k$, so your $\phi(ij)=\phi(i)\phi(j)$ would suffice.

Answer 2

I found a solution online.

$Q_8$ can be expressed as $\{i,j \; | i^4=1, i^2=j^2, ji=ij^{-1} \} $.

So, we need to prove that $\varphi(i)$ and $\varphi(j)$ satisfy these relations. Since they do, we can conclude $\varphi$ is a homomorphism from $Q_8$ to $GL_n(\mathbb{C})$.