Let $G$ be $SL_2(\mathbb{F}_5)$ i.e. the special linear group of $2\times 2$ matrices $\mathbb{F}_5$.

Find a subgroup of $G$ isomorphic to $Q_8$, and an element of order $3$ normalizing it in $G$.

I found a subgroup $Q_8$, but it's really random - is there an obvious simple one?

I'm not sure what the element of order three normalizing it in G means - I originally thought it meant that $Q_8$ with the new element (i.e. a new subgroup of order $24$, which i think is the last part of the question) makes a normal subgroup in G, but the only normal subgroup in $G$ is $\{I,-I\}$.

I've obviously misunderstood the question...

what exactly does this new element of order three do??

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    $\begingroup$ There are three questions here. Do you see how to do any of them? Do you see how the last one follows from the two first ones, for example? $\endgroup$ – Tobias Kildetoft Dec 27 '13 at 16:50
  • $\begingroup$ Dear user, To say that $g \in G$ normalizes the subgroup $H$ of $G$ is to say that $gHg^{-1} = H$, i.e. the subgroup of $G$ you get by conjugating every element of $H$ by $g$ is equal to the original subgroup $H$. Regards, $\endgroup$ – Matt E Dec 28 '13 at 2:34
  • $\begingroup$ what is your subgroup which is isomorphic to $Q_8$ $\endgroup$ – user87543 Dec 28 '13 at 3:06
  • $\begingroup$ I think I got it now! I can let the normalizing element be 2I. Thanks! $\endgroup$ – user117908 Dec 28 '13 at 3:23
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    $\begingroup$ There is a $Q_8$ subgroup that's not hard to describe - think of diagonal and antidiagonal matrices. I don't know whether there is any easy approach to finding the normalizeing element of order $3$. $\endgroup$ – Derek Holt Dec 28 '13 at 13:57

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