Generators of $SL_2({\bf F}_3)$

Question

Question : Show that the followings are generators of $SL_2({\bf F}_3)$ :

$$ a=\left(\begin{array}{cc} 1&1\\ 0&1 \end{array}\right),\ b =\left(\begin{array}{cc} 1&0\\ 1&1 \end{array}\right)$$

This is an exercise of Dummit and Foote.

My solution : $H= SL_2({\bf F}_3)$ has index $2$ in $GL_2({\bf F}_3)$ so that it has order $24$. (In problem, order of $GL_2({\bf F}_3)$ is given)

Since $a$ has order $3$ we must show that $H$ does not have order $12$. Assume that $H$ is a non-abelian group of order $12$

By computation we can show that $a^3=b^3=1$, $ab$ has order $4$ and $ba=babb^{-1}$ is not in $\langle ab\rangle$ so that this group is not normal.

Hence $$n_3=4,\ n_2>1$$

By classification of group of order $12$, the above case cannot happen. So we complete the proof.

Question : My proof is regular ? I have a doubt in computation proof. If you have a proof which is less computing, please tell me. Thank you.

score 3 · Accepted Answer · answered Jan 03 '14 at 03:05

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Given the size of the group, you can just compute the 12 matrices in $SL(2, F_3)$ as products of your two generators. It beats thinking by a long shot.

answered Jan 03 '14 at 03:05

Igor Rivin

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score 1 · Answer 2 · answered Jan 03 '14 at 04:24

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I think a and b are just the elementary transformations. So you can reduce every matrix to identity only by these two transformations, which means a b are generators.

answered Jan 03 '14 at 04:24

Wei Zhou

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Thank you for your reply. But I cannot understand. – HK Lee Jan 03 '14 at 07:44
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If you know the connection of elementary matrix and elementary transformation, you can understand what I say easily. – Wei Zhou Jan 03 '14 at 07:58
I see (You mean that for instance $a$ and $a^2$ are ${\bf row \ operation}$s). Thank you. – HK Lee Jan 03 '14 at 08:04
yes, you are right. – Wei Zhou Jan 03 '14 at 08:12

Generators of $SL_2({\bf F}_3)$

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