When is PSU(2,q^2) = PSL(2,q) ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T04:04:17Z http://mathoverflow.net/feeds/question/120239 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120239/when-is-psu2-q2-psl2-q When is PSU(2,q^2) = PSL(2,q) ? Will Chen 2013-01-29T18:08:22Z 2013-01-30T12:11:56Z <p>The context for this question is from page 284 - 287 of Berger's paper: <a href="http://pdn.sciencedirect.com/science?_ob=MiamiImageURL&amp;_cid=272332&amp;_user=209810&amp;_pii=S0021869398976785&amp;_check=y&amp;_origin=article&amp;_zone=toolbar&amp;_coverDate=1999--01&amp;view=c&amp;originContentFamily=serial&amp;wchp=dGLbVlS-zSkzS&amp;md5=2dd8e0d714d264cf7c4acdd9ec58ac84&amp;pid=1-s2.0-S0021869398976785-main.pdf" rel="nofollow">http://pdn.sciencedirect.com/science?_ob=MiamiImageURL&amp;_cid=272332&amp;_user=209810&amp;_pii=S0021869398976785&amp;_check=y&amp;_origin=article&amp;_zone=toolbar&amp;_coverDate=1999--01&amp;view=c&amp;originContentFamily=serial&amp;wchp=dGLbVlS-zSkzS&amp;md5=2dd8e0d714d264cf7c4acdd9ec58ac84&amp;pid=1-s2.0-S0021869398976785-main.pdf</a></p> <p>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 $k = \sharp\langle -\zeta\rangle$."</p> <p>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 $\text{PSL}_2(\mathbb{F}_q)$.</p> <p>Is this generally true?</p> <p>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.</p> <p>Where might I find a good book on unitary matrices over finite fields?</p> <p>thanks,</p> <ul> <li>will</li> </ul> http://mathoverflow.net/questions/120239/when-is-psu2-q2-psl2-q/120267#120267 Answer by Peter Mueller for When is PSU(2,q^2) = PSL(2,q) ? Peter Mueller 2013-01-29T22:39:46Z 2013-01-29T22:39:46Z <p>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 <em>Endliche Gruppen</em>. I would expect that it is also in the third volume of <em>The Classification of the Finite Simple Groups</em> by Gorenstein-Lyons-Solomon, but I don't have the volume at hand right now.</p> http://mathoverflow.net/questions/120239/when-is-psu2-q2-psl2-q/120290#120290 Answer by Derek Holt for When is PSU(2,q^2) = PSL(2,q) ? Derek Holt 2013-01-30T09:14:24Z 2013-01-30T12:11:56Z <p>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 $\phi\gamma$.</p> <p>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 <code>$\left( \begin{array}{rr}0&amp;1\\ -1&amp;0\end{array} \right)$</code>. 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 <code>$g = \left( \begin{array}{rr}a&amp;b\\ c&amp;d\end{array} \right)$</code>, where $b = -t^qa^q$ and $d= -t^qc^q$ for some field element $t$ with $t^{q+1} = -1$.</p>