# Show $(M,+)$ is isomorphic to $(\mathbb{C},+)$

Let, $M=\{ \begin{pmatrix}a&-b\\b&a\end{pmatrix} :a,b\in \mathbb{R}\}$, show $(H,+)$ is isomorphic as a binary structure to $(\mathbb{C},+)$

Define $f : M\rightarrow \mathbb{C}$ by $f\begin{pmatrix}a&-b\\b&a\end{pmatrix} = a+bi$

Let $a,b,c,d\in \mathbb{R}$

$1-1$:

Suppose $f\begin{pmatrix}a&-b\\b&a\end{pmatrix}=f\begin{pmatrix}c&-d\\d&c\end{pmatrix}$, then $a+bi =c+di$, thus $a=c$ and $b=d$, so $f$ is one to one.

Onto:

Let $a+bi\in \mathbb{C}$ , then $\begin{pmatrix}a&-b\\b&a\end{pmatrix}\in M$, so $f\begin{pmatrix}a&-b\\b&a\end{pmatrix}=a+bi$, thus $f$ is onto.

Homomorphic:

\begin{align*} &f(\begin{pmatrix}a&-b\\b&a\end{pmatrix} + \begin{pmatrix}c&-d\\d&c\end{pmatrix})\\ =&f(\begin{pmatrix}a+c&-(b+d)\\b+d&a+c\end{pmatrix})\\ = &(a+c)+(b+d)i\\ = &f(\begin{pmatrix}a&-b\\b&a\end{pmatrix}) +f(\begin{pmatrix}c&-d\\d&c\end{pmatrix})\end{align*}

I don't think I show $f: M\rightarrow \mathbb{C}$ is $1-1$ and onto correctly. can any give me a hit to show $f: M\rightarrow \mathbb{C}$ is $1-1$ and onto? thanks!

• This all looks fine to me. The more interesting piece of the puzzle - which wasn't necessarily part of this problem - is to show that $M$ and $\mathbb{C}$ are isomorphic with respect to multiplication... – Steven Stadnicki Apr 8 '15 at 4:48
• $(M,+)\cong \mathbb R^2 \cong (\mathbb C, +)$ – Matthew Levy Apr 8 '15 at 5:30
• That is correct, but I have two points. First, the $H$ in the statement should be $M$, and the second is that, for the sake of completeness of your answer, you need also check that it preserves the identity element. – Math137 Apr 8 '15 at 9:19
• @Math137 the identity is the zero matrix, do I need to show M (0) map to C or there exist an identity , 0 matrix, that M+0=M? – Simple Apr 8 '15 at 16:31
• @Simple you need to show that your map takes the zero matrix to zero of $\mathbb{C}$, which is not hard but necessary, because it is a part of the structure of the group. – Math137 Apr 8 '15 at 16:33