Group $G$ is the set of all matrices that are real and invertible with this form: $\begin{pmatrix}m&-n\\ n&m\end{pmatrix}$
Show that $f : C^∗ → G$ defined by $f(m + ni) = \begin{pmatrix}m&-n\\ n&m\end{pmatrix}$ is an isomorphism.
My solution:
$C^*$ is complex group under multiplication. So, to show $f$ is isomorphism, I need to show $f$ is bijective and that $f(ab) = f(a)f(b)$. Here I get stuck a bit...
$f((a+bi)(x+iy)) = f(ax+xbi+ayi+iybi) = f(ax+(ab+ay+ybi)i) = \begin{pmatrix}ax&-\left(ab+ay+ybi\right)\\ \:ab+ay+ybi&ax\end{pmatrix}\:$ And, $f(a+bi)f(x+iy) = \begin{pmatrix}a&-b\\ \:b&a\end{pmatrix}\: \begin{pmatrix}x&-y\\ \:\:y&x\end{pmatrix}\: = \begin{pmatrix}ax-by&-ay-bx\\ bx+ay&-by+ax\end{pmatrix}$
...yea, so I'm having a hard time with showing $f(ab) = f(a)f(b)$ and that $f$ is bijective.
First, $f(a + ib) = f(m + in)$ if and only if $\begin{pmatrix} a && -b \\ b && a\end{pmatrix} =\begin{pmatrix} m && -n \\ n && m\end{pmatrix} $ if and only if $a = m$ and $b = n$, then $f$ is an injective function.
$f$ is surjective, well $\begin{pmatrix} a && -b \\ b && a\end{pmatrix} = f(a + ib)$ for all $a,b \in C^{\ast}$
Now, $f((a + ib)(n + im)) = f(an - bm + i(am + nb)) = \begin{pmatrix} an - bm && -am - bn \\ am + bn && an - bm\end{pmatrix} $
And $f(a + ib)f(n + im) = \begin{pmatrix} a && -b \\ b && a\end{pmatrix} \begin{pmatrix} n && -m \\ m && n\end{pmatrix} = \begin{pmatrix} an - bm && -am - bn \\ bn + am && -bm + an\end{pmatrix} $
Thus $f((a+ib)(n+im)) = f(a+ib)f(n+im)$, $f$ is a bijective homomorphims. Is an isomorphims.