Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it 'expected' that PSL(2,p) contains a cyclic subgroup of order $\tfrac{p+1}{2}$? I read such an element exists and generates a "nonsplit torus", but I'm not sure what that means.

Is there an easy explanation why such an element should exist and have a cycle structure $\tfrac{p+1}{2},\tfrac{p+1}{2}$ in its action on the projective line?

(Edit: On a sidenote, it is on the other hand obvious that PSL(2,p) has cyclic subgroups of order (p-1)/2 and p, generated by $(\begin{smallmatrix}r&0\\0&\tfrac{1}{r}\end{smallmatrix})$ and $(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix})$ where r generates $GF(p)^*$ )

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Elements of $\text{SL}_2(\mathbb{F}_p)$ have eigenvalues in $\mathbb{F}_{p^2}$. The multiplicative group of $\mathbb{F}_{p^2}$ is cyclic of order $p^2 - 1$ generated by some element $g$, so it follows that $\text{GL}_2(\mathbb{F}_p)$ has a cyclic subgroup of order $p^2 - 1$ (just write out the matrix of multiplication by $g$). The determinant of this matrix is some element of $\mathbb{F}_p^{\ast}$, so $g^{p-1}$ has determinant $1$ and order $p+1$. Finally, the image of $g^{p-1}$ in $\text{PSL}_2(\mathbb{F}_p)$ has order $\frac{p+1}{2}$ since $g^{ \frac{p^2 - 1}{2} } = -1$. You can work out the cycle structure from here.

As for what a nonsplit torus is, the terminology comes from algebraic geometry and I can't help you there.

share|improve this answer
    
Thank you! That was a helpful (and quick) reply. –  Myself Jan 3 '11 at 17:12
    
If it helps, the "split torus" is the cyclic group of order (p-1)/gcd(p,2). You can think of it like polynomials. x^(p-1)-1 splits over Fp, so it gives you the split torus, but (x^(p^2-1)-1)/(x^(p-1)-1) does not split, and gives you the non-split torus. This works for p a prime power too, though the "order p" subgroup is no longer cyclic. It is called "unipotent". –  Jack Schmidt Jan 3 '11 at 17:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.