# When is PSU(2,q^2) = PSL(2,q) ?

Particularly, in his assumption at the top of page 287, he says that "From now on, assume that our map $\pi_\mathfrak{p}$ surjects onto $\text{PU}_2(\zeta,\mathcal{O}_K/\mathfrak{p})\cong \text{PSL}_2(\mathbb{F}_q)$, that $q$ is odd, and that $(6,k) = 1$, where $k = \sharp\langle -\zeta\rangle$."

I'm guessing that he's assuming the conditions in proposition 2 (from the previous page) to be true, so that $\pi_\mathfrak{p}$ surjects onto $\text{PU}_2(\zeta,\mathcal{O}_K/\mathfrak{p}) = \text{PSU}_2(\mathcal{O}_K/\mathfrak{p})$, and that he's claiming that the latter group is isomorphic to $\text{PSL}_2(\mathbb{F}_q)$.

Is this generally true?

Also, on page 284, where he gives the matrix $H$ for the hermitian form, he claims that $H\in GL_{n-1}(\mathbb{Z}[t,t^{-1}])$, but the matrix he gives obviously does not lie in that group.

Where might I find a good book on unitary matrices over finite fields?

thanks,

• will
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There are various questions mixed into your text, but it would help to clarify what you actually mean by $\mathrm{PSU}_2$ in the header. From the finite group perspective, there is only one family of type $A_1$ simple groups, usually denoted $\mathrm{PSL}_2$. (Though in higher ranks there are different split and non-split simple groups of type $A_\ell$.) – Jim Humphreys Jan 29 '13 at 20:23

Here is a more bare hands explanation. Let $\phi$ be the field automorphism of ${\rm SL}_n(q^2)$ that acts by applying $x \mapsto x^q$ to the matrix entries. Let $\gamma$ be the graph automorphism that maps matrices $A$ to their inverse-tranpose $A^{- \mathrm{T}}$. Then ${\rm SL}_n(q)$ is the subgroup of ${\rm SL}_n(q^2)$ that is centralized by $\phi$, whereas the group ${\rm SU}_n(q^2)$ (which is confusingly often denoted by ${\rm SU}_n(q)$) that fixes the identity matrix as unitary form is the subgroup of ${\rm SL}_n(q^2)$ that is centralized by $\phi\gamma$.
The automorphism $\gamma$ is outer for $n>2$, but when $n=2$ it is inner and acts in the same way as conjugation by the matrix $\left( \begin{array}{rr}0&1\\ -1&0\end{array} \right)$. It turns out in this case that $\phi$ and $\phi\gamma$ are conjugate in the automorphism group of ${\rm SL}_2(q^2)$ by (the projective image of) an element $g \in {\rm GL}_2(q^2)$, and hence that ${\rm SL}_2(q)$ is conjugate to ${\rm SU}_2(q^2)$ in ${\rm GL}_2(q^2)$. With a bit of calculation on the back of an envelope, we find that $g = \left( \begin{array}{rr}a&b\\ c&d\end{array} \right)$, where $b = -t^qa^q$ and $d= -t^qc^q$ for some field element $t$ with $t^{q+1} = -1$.
What do you mean "centralized by $\phi$"? Do you mean "fixed by $\phi$"? – William Chen Jan 30 '13 at 21:24
I mean the subgroup consisting of those elements that are fixed by $\phi$. I prefer to call it centralized because that makes it clear that I mean fixed element-wise rather than fixed as a set. – Derek Holt Jan 30 '13 at 22:31
It is a fairly standard result that $SU(2,q^2)$ and $SL(2,q)$ are isomorphic, see e.g. II.8.8. in Huppert's Endliche Gruppen. I would expect that it is also in the third volume of The Classification of the Finite Simple Groups by Gorenstein-Lyons-Solomon, but I don't have the volume at hand right now.
@Peter: In the G-L-S book, pages 68-69 summarize special linear and special unitary groups, noting the degenerate 2-dimensional case $m=1$ (first full paragraph on page 69). Labels get complicated for the classical groups, but I guess the moral is that in dimension 2 there is no real need to worry about unitary matrices over finite fields. – Jim Humphreys Jan 30 '13 at 1:31