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What are the finite subgroups of $GL_2(\mathbb{Z})$?

I would quite like to know what the matrices which generate the subgroups are.

I know that this group has an index two subgroup which is isomorphic to $\langle x, y; x^6, y^4, x^3=y^2\rangle$ but

  • a) I cannot remember facts about free products with amalgamation for more than $5$ minutes
  • b) I do not know which matrix the $x$-generator corresponds to (or, I suppose, the $y$-generator, but I can easily find elements of order $4$...$6$ is more elusive).

Note: I have edited this answer as I got mixed up with $SL$ and $GL$ for some reason...

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1 Answer

From Trees by Serre (chapter 3, section 4.3, corollary of theorem 8):

Theorem: Let $G$ be an amalgam $G_1 \underset{A}{\ast} G_2$ of two groups. Every finite subgroup of $G$ is contained in a conjugate of $G_1$ or $G_2$.

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If I am not mistaken, it is $\mathbb{SL}_2(\mathbb{Z})$ which is an amalgam: it is isomorphic to $C_4 *_{C_2} C_6$. So a little more work seems to be needed here. –  Pete L. Clark Feb 5 '13 at 16:01
    
Phoo! I've written down the presentation for $SL_2(\mathbb{Z})$ not $GL_2(\mathbb{Z})$. There is definately a copy of $C_2\times C_2$ in $GL_2(\mathbb{Z})$ (take the four matrices with diagonal entries of absolute value $1$ and $0$s everywhere else.) –  user1729 Feb 5 '13 at 16:05
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@user1729: No worries. I just noticed that the presentation the OP has written down is also for $\operatorname{SL}_2(\mathbb{Z})$. So maybe it's safest to give both answers. Anyway, to get from $GL_2$ back to $SL_2$ just multiply everything by the diagonal matrix with entries $(1,-1)$: no biggie. –  Pete L. Clark Feb 5 '13 at 16:07
    
@PeteL.Clark: Sure, okay, but what are the matrices of order $4$ and $6$? I mean, I could guess a candidate for the one of order $4$, but I cannot find any of order $6$. –  user1729 Feb 5 '13 at 16:09
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A matrix of order 3 is $A=\left(\begin{array}{rr}0&1\\-1&-1\end{array}\right)$ so $-A$ has order 6. You can take $B=\left(\begin{array}{rr}0&1\\-1&0\end{array}\right)$ as your matrix of order 4. Then $A$ and $B$ generate ${\rm SL}_2(\mathbb{Z})$. –  Derek Holt Feb 5 '13 at 17:21
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