Prove that $\mathbb{C}^*$ is isomorphic to the subgroup of $GL_2(\mathbb{R})$ consisting of matrices
of the form
$$\begin{bmatrix}
a & b\\
-b & a
\end{bmatrix}$$
I showed that $\mathbb{C}^*$ and $GL_2(\mathbb{R})$ are homomorphic by proving that $\Phi(z$ o $w)$ = $\Phi (z)$ . $\Phi (w)$ where $z,w$ are elements of $\mathbb{C}^*$.
One-to one? I have shown that for any two complex $z$ and $w$ when $$\Phi (z) =\Phi (w) => z=w$$
Onto?
Let
$$\begin{bmatrix}
m & n\\
p & q
\end{bmatrix} \in GL_2(\mathbb{R})$$
and let $$\Phi (a+bi) = \begin{bmatrix} m & n \\ p & q\end {bmatrix}.$$
Then $$\begin{bmatrix}
a & b\\
-b & a
\end{bmatrix} = \begin{bmatrix} m & n \\ p & q\end {bmatrix}$$
BUT then what can be said about that to prove that the relation is onto? i am a bit stuck here.
Best Answer
Your map $\Phi$ isn't to $GL_2(\Bbb R)$ itself, but rather the subgroup $G = \left\{\pmatrix{a & b \\ -b & a}: a, b \in \Bbb R\right\}$. The instructions aren't incredibly explicit, but it is very important here that $\Phi$ has codomain $G$, rather than all of $GL_2(\Bbb R)$.
When you pick a matrix $M =\pmatrix{m&n\\p&q}$, you're not necessarily picking something in $G$, the codomain of $\Phi$. It may very well be the case that there's no $z \in \Bbb C^*$ such that $\Phi(z) = M$; this happens if and only if $q = m$ and $p = -n$. So if you pick the $M$ above as your generic matrix and try to find some $z \in \Bbb C^*$ so that $\Phi(z) = M$, you had better not be able to show that $\Phi$ is onto $GL_2(\Bbb R)$ -- it's not!
Instead, to show that $\Phi$ is surjective ("onto"), pick something in the codomain of $\Phi$, something in $G$.
So pick a generic matrix $M = \pmatrix{m&n\\-n&m} \in G$, and find the corresponding $z \in \Bbb C^*$ with $\Phi(z) = M$.
In actuality, defining $\Phi$ as above but making its codomain all of $GL_2(\Bbb R)$ instead of just $G$ isn't a huge problem. It's a general fact that for any group homomorphism $\varphi: H \to H'$, if $\varphi$ is injective, then $H$ is isomorphic to the image $\operatorname{im}(\varphi) = \{\varphi(h) : h \in H\}$ (this is a consequence of the first isomorphism theorem).
Since you've shown that $\Phi$ is an injective homomorphism from $\Bbb C^*$, then $\Bbb C^*$ is necessarily isomorphic to $\operatorname{im}(\Phi)$, which is exactly the $G$ defined in the problem statement. It's just that, unless the codomain of $\Phi$ is exactly $\operatorname{im}(\Phi)$, you won't be able to show that $\Phi$ is surjective.