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Show that group of all real matrices of form $$ \begin{bmatrix} x & y\\ -y & x \end{bmatrix} , \qquad (x,y) \ne (0,0) $$ is isomorphic with/to $\mathbb C \setminus \left\{{0}\right\}$ under complex multiplication?

I know two ways to show isomorphism: 1) finding a homomorphic function 2) writing the multiplication table and comparing.

I think the latter is possible for finite groups.

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    $\begingroup$ The condition $xy \ne 0$ is wrong. You mean $x$, $y$ not both $0$. As it is, you don't have a group. $\endgroup$ – Robert Israel Sep 29 '13 at 7:28
  • $\begingroup$ What is the identity element of the group? If it is $I_2$, then $y=0$. $\endgroup$ – Falang Sep 29 '13 at 7:31
  • $\begingroup$ @RobertIsrael I checked all conditions of groups and all of them held. Why it is not a group? $\endgroup$ – Mahdi Khosravi Sep 29 '13 at 8:01
  • $\begingroup$ It doesn't have an identity and isn't closed under multiplication. $\endgroup$ – Robert Israel Sep 29 '13 at 8:05
  • $\begingroup$ Mahdi, @RobertIsrael already told you in his first comment how to fix the assumptions. You want to exclude the zero matrix only. $\endgroup$ – Andreas Caranti Sep 29 '13 at 8:21
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Let $G$ be the group of those matrices with usual matrix multiplication. Define

$$f:G\to \Bbb C\setminus \{0\}$$ $$f\big(\begin{bmatrix} x & y\\ -y & x \end{bmatrix}\big)=x+iy$$ Show that it is an isomorphism.

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  • $\begingroup$ I do not want this answer be deleted. $\endgroup$ – user79193 Sep 30 '13 at 14:47
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$\newcommand{\C}{\mathbb{C}}\newcommand{\R}{\mathbb{R}}$What's behind this exercise is the following.

Consider $\C$ as a vector space over $\R$, with basis $1, i$. For each $x + i y \in \C$, with $x, y \in \R$, consider the map $$ \C \to \C \qquad z \mapsto z \cdot (x + i y). $$ This map is $\R$-linear, and its matrix with respect to the basis $1, i$ is precisely $$ \begin{bmatrix} x & y\\ -y & x \end{bmatrix} $$

See if you can get on from here.

Note The matrix is the one above if you consider row vectors. If you consider column vectors, then take the map $z \mapsto z \cdot (x - i y)$, which is still $\R$-linear.

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  • $\begingroup$ Thanks for your answer. But I am not familiar with some of terms you've used. Isn't there any simpler solution? $\endgroup$ – Mahdi Khosravi Sep 29 '13 at 8:27
  • $\begingroup$ @MahdiKhosravi, you should simply check that multiplying the two complex numbers $x_{k} + i y_{k}$, for $k = 1, 2$, is the same as multiplying the two matrices \begin{bmatrix} x_{k} & y_{k}\\ -y_{k} & x_{k} \end{bmatrix} $\endgroup$ – Andreas Caranti Sep 29 '13 at 8:41
  • $\begingroup$ @MahdiKhosravi: All answers from this user are more preferable for mine, Mahdi. $\endgroup$ – mrs Oct 16 '13 at 10:14
  • $\begingroup$ @BabakS. Thanks for your comment. $\endgroup$ – Mahdi Khosravi Oct 16 '13 at 11:31
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You should never need to write out a multiplication table in order to prove that two groups are isomorphic, unless the groups given are actually defined by their multiplication table.

In this case, the groups are infinite anyway, so you need to find a homomorphic function and show that it's homomorphic and a bijection.

Some hints to help you find such a function:

  • Consider the case $y=0$. How do the matrices $\begin{pmatrix}x&0\\0&x\end{pmatrix}$ behave under multiplication? What does that remind you of in the complex numbers?

  • What is $\begin{pmatrix}0&-1\\1&0\end{pmatrix}^2$? What does that remind you of?

  • Using these, what seems like the most natural map from matrices of the form $\begin{pmatrix}x&-y\\y&x\end{pmatrix}$ ($x,y\ne0$) to complex numbers of the form $a+ib$ ($a,b\ne0$)? Can you prove that that map is a homomorphism and a bijection?

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  • $\begingroup$ Thanks for your answer. But I didn't get what map is the most natural. I am a beginner! $\endgroup$ – Mahdi Khosravi Sep 29 '13 at 9:22
  • $\begingroup$ @Mahdi It's very important that you complete the first two bullet points, before moving on to the third. Have you done the first two? What were your answers? $\endgroup$ – John Gowers Sep 29 '13 at 17:02
  • $\begingroup$ I don't know I am right or not. First bullet says when $y=0$ the multiplication is similar to the real case. And the second one says that when $x=0$, odd powers are in one form (similar to itself) and even powers are in another form (the form when $y=0$) $\endgroup$ – Mahdi Khosravi Sep 29 '13 at 17:08
  • $\begingroup$ @Mahdi Does that remind you of complex numbers in any way? $\endgroup$ – John Gowers Sep 29 '13 at 17:10
  • $\begingroup$ You mean $x$ represents real part and $y$ represents imaginary part? $\endgroup$ – Mahdi Khosravi Sep 29 '13 at 17:18

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