Ok, I believe this is a relatively basic question, but I couldn’t figure out what I’m doing wrong. What I’d like to prove is the following result:
If an element $T$ in $SL(2, \mathbb{Z})$ has finite order $n$, then $n = 1, 2, 3, 4$ or $6$
I have shown before that $PSL(2, \mathbb{Z}) \simeq \mathbb{Z}_2*\mathbb{Z}_3$, and I’d really like to use Kurosh’s Subgroup Theorem. One consequence of such theorem is that any element of finite order has to be conjugate to an element of finite order in either $\mathbb{Z}_2$ or $\mathbb{Z}_3$.
I can thus see why orders $1,2,3$ are all possible. But conjugation is an automorphism - therefore, the order of the conjugate is the same as that of the original element. So I don’t see why $4$ and $6$ are possible, as neither group has any element of order above $3$…
I can prove this fact otherwise, but I’ve just studied Kurosh’s Theorem, and am specifically trying to apply it. I have not yet studied amalgams, though, so I would like to use it directly in the (unamalgamated) free product above.
What am I doing wrong?
Thanks in advance!
EDIT: @Derek Holt pointed out that the real result is in $SL(2, \mathbb{Z})$. I will try to lift it from the decomposition in $PSL$