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September 14th, 2017, 02:08 PM   #1
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automophism

I want to show that there exists an automorphism $\phi$ of $Q_8$ such that $\phi(\sigma)=\nu$ and $\phi(\nu)=\sigma\nu$.


Where $\sigma=\begin{pmatrix}e^{\pi \frac{i}{2}}&0\\0&e^{-\pi \frac{i}{2}}\end{pmatrix}$ and $\nu=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$

Thanks for your help in advance

Last edited by mona123; September 14th, 2017 at 02:19 PM.
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September 14th, 2017, 02:15 PM   #2
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What work have you done on the problem? Where are you stuck? Have you written down the exact definition of each technical term in the problem? Have you written down exactly what it is you're being asked to show? These steps are essential in every problem you will encounter at this level of math.
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September 14th, 2017, 02:28 PM   #3
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I have noticed only that $\phi(\nu)=\sigma\phi(\sigma)$ but i can't answer the problem, please help me i will be grateful if you do so.

Last edited by greg1313; September 14th, 2017 at 04:49 PM.
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September 29th, 2017, 05:19 AM   #4
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The first thing I would do is calculate that
$\displaystyle \sigma\nu= \begin{pmatrix}e^{\pi\frac{i}{2}} & 0 \\ 0 & e^{-\pi\frac{i}{2}}\end{pmatrix}\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}= \begin{pmatrix}0 & e^{\pi\frac{i}{2}} \\ -e^{-\pi\frac{i}{2}} & 0 \end{pmatrix}$.

So you want $\displaystyle \phi$ that maps $\displaystyle \begin{pmatrix}e^{\pi\frac{i}{2}} & 0 \\ 0 & e^{-\pi\frac{i}{2}}\end{pmatrix}$ to $\displaystyle \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$ and maps $\displaystyle \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$ to $\displaystyle \begin{pmatrix}0 & e^{\pi\frac{i}{2}} \\ -e^{-\pi\frac{i}{2}} & 0 \end{pmatrix}$.
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