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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!

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    $\begingroup$ 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... $\endgroup$ – Steven Stadnicki Apr 8 '15 at 4:48
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    $\begingroup$ $(M,+)\cong \mathbb R^2 \cong (\mathbb C, +)$ $\endgroup$ – Matthew Levy Apr 8 '15 at 5:30
  • $\begingroup$ 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. $\endgroup$ – Math137 Apr 8 '15 at 9:19
  • $\begingroup$ @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? $\endgroup$ – Simple Apr 8 '15 at 16:31
  • $\begingroup$ @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. $\endgroup$ – Math137 Apr 8 '15 at 16:33
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Responding to the request from @Julian and to move this question to answered question list, I will try to sum up what has been mentioned in the comments.

Yes, your answer is correct.

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protected by user26857 Sep 7 '15 at 14:13

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