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Let $G$ be the group of all nonzero complex numbers under multiplication and let $\bar{G}$ be the group of all real $2\times 2$ matrices of the form $\begin{pmatrix} a & b \\ -b & a \end{pmatrix},$ where not both $a$ and $b$ are $0$, under matrix multiplication. Show that $G$ and $\bar{G}$ are isomorphic by exhibiting an isomorphism of $G$ onto $\bar{G}.$

If I let $\bar{G} = \begin{pmatrix} a & b \\ -b & a \end{pmatrix},$ then I define a mapping by: $\varphi: \bar{G} \to G,$ thus $\varphi\begin{pmatrix} a & b \\ -b & a \end{pmatrix} = a+bi$.

Now let $A=\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ and $B = \begin{pmatrix} p & q \\ -q & p \end{pmatrix},$ where $A, B \in \bar{G}.$ Then: $$\varphi (AB) = \varphi\left[\begin{pmatrix} a & b \\ -b & a \end{pmatrix}\begin{pmatrix} p & q \\ -q & ap\end{pmatrix}\right] = \varphi\left[\begin{pmatrix} ap-bq & aq+bp \\ -(aq+bp) & ap-bq \end{pmatrix}\right]=(ap-bq)+(aq+bp)i.$$

How can I complete this proof?

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(i) Your last line is not yet complete; it should read $$\eqalign{\varphi (AB) &= \varphi\left(\left[\matrix{ a & b \cr -b & a \cr}\right]\cdot\left[\matrix{ p & q \cr -q & p\cr}\right]\right)\cr &= \varphi\left(\left[\matrix{ ap-bq & aq+bp \cr -(aq+bp) & ap-bq \cr}\right]\right)\cr&=(ap-bq)+(aq+bp)i\cr &=(a+bi)\cdot(p+qi)\cr &=\varphi(A)\cdot\varphi(B)\ .\cr}$$ (ii) You should note that the map $\phi$ is bijective, even if it's more or less obvious.

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  • $\begingroup$ Why did someone downvote this answer? And it seems that someone downvoted my question too? $\endgroup$ – Lays Oct 9 '13 at 21:06
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you need to check that your function is bijective. So far you now that $\varphi$ is a homomorphism. Now check that if $\varphi(A) = \varphi(B) \implies A = B$. Ths is obvious therefore it is obvious that your function is injective. For surjective, find an element in $A \in \bar{G}$ such that $\varphi(A) = z \in G$. Then you would have an isomorphism.

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  • $\begingroup$ For checking injective, I would suggest only looking at the kernel, rather than at arbitrary elements. That is slightly simpler. $\endgroup$ – Tobias Kildetoft Oct 9 '13 at 8:45
  • $\begingroup$ And don't forget to check that $\varphi(Id)=1$. $\endgroup$ – A. Bellmunt Oct 9 '13 at 8:46
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    $\begingroup$ @A.Bellmunt That follows automatically from the map being a homomorphism. $\endgroup$ – Tobias Kildetoft Oct 9 '13 at 8:49
  • $\begingroup$ @TobiasKildetoft True enough. Can you assert that even if the homomorphism is not surjective? Sorry, I think I'm a bit unfocused today. $\endgroup$ – A. Bellmunt Oct 9 '13 at 9:23
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    $\begingroup$ @A.Bellmunt Yes, for groups this is true. For monoids, it can fail (the image of the identity will be an identity for the image, and a group cannot have "local" identity elements other than the actual identity element) $\endgroup$ – Tobias Kildetoft Oct 9 '13 at 9:26

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