Let $SL_2 (q)$ be group of all $2\times 2$ invertible matrices with unit determinant and $PGL_2(q)$ is quotient group $GL_2(q)/\{\text{scalar matrices over}\ q\}$.
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$\begingroup$ what is a root systems there is an simple answer? $\endgroup$– SinaCommented Jul 10, 2012 at 12:57
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1$\begingroup$ Order of $SL_2(q)$ equals $q(q^2-1)$, order of $PGL_2(q)$ is $q(q^2-1)(q+1)$. $\endgroup$– Andrei SmolenskyCommented Jul 10, 2012 at 13:09
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3$\begingroup$ They have the same order. $PGL_2(q)=GL_2(q)/\{\lambda I, \lambda\in F_q-\{0\}\}$ where I is the identity matrix $\endgroup$– SinaCommented Jul 10, 2012 at 13:21
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4$\begingroup$ Dear Sina, It's not clear that the answer you have accepted actually answers your question. Indeed, as Jim Humphreys points out in his answer, the groups are not isomorphic if $q$ is odd (the first has a non-trivial centre while the second does not), but are isomorphic if $q$ is a power of $2$. Regards, $\endgroup$– EmertonCommented Jul 10, 2012 at 14:48
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1$\begingroup$ Oops! I just thought that what I wrote above is wrong. The lesson I learned today: never try to reproduce a formula you believe to remember without checking. $\endgroup$– Andrei SmolenskyCommented Jul 10, 2012 at 19:11
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3 Answers
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The question itself is natural, but it's fairly elementary and has a clearcut answer in the literature on finite simple groups including the series of books by Gorenstein-Lyons-Solomon (and for small order groups the Atlas). It's easiest to understand what is going on from the algebraic group viewpoint, summarized with references in Section 1.1 of my 2006 LMS Lecture Note volume Modular Representations of Finite Groups of Lie Type. Here Lang's theorem is crucial. It shows that whenever you have an isogeny (algebraic group epimorphism with finite kernel) from one connected algebraic group onto another over a finite field of $q$ elements, the corresponding groups of rational points over $\mathbb{F}_q$ have the same order.
In your case, start with the natural map from a general linear group to the quotient by scalars, which restricts to an isogeny $\mathrm{SL}_2 \rightarrow \mathrm{PGL}_2$. Thus the two finite groups do have the same order for any $q$, even though the original map fails to be an algebraic group isomorphism. When $q$ is even, however, the finite groups are in fact isomorphic. But when $q$ is odd, the group on the left has a nontrivial center and the group on the right doesn't.
answered Jul 10, 2012 at 14:03
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Not quite, $PGL(2, F_q) \cong PSL(2, F_q) \rtimes F_q^\times/ (F_q^{\times})^2$.
answered Jul 10, 2012 at 14:33
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2$\begingroup$ But when $q$ is a power of $2,$ every non-zero element of the field is a square, and the ${\rm PSL}$ on the right is an ${\rm SL}.$ $\endgroup$ Commented Jul 10, 2012 at 15:50
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1$\begingroup$ I do not understand. Is there a mistake, of are you saying the OP is correct in this special case? $F_q^\times$ modulo the squares is a trivial group, when $2$ divides q. $\endgroup$ Commented Jul 10, 2012 at 16:41
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1$\begingroup$ Dear Mrc, I think that Geoff Robinson is objecting to your statement "not quite". If $q$ is a power of $2$ then in fact the general isomorphism does reduce to an isomorphism between $PGL_2$ and $SL_2$. Regards, $\endgroup$– EmertonCommented Jul 10, 2012 at 17:25
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$\begingroup$ I was scared for a second here;) $\endgroup$ Commented Jul 10, 2012 at 18:39
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$SL_2$ and $PGL_2$, seen as linear algebraic groups, have different root data. See Milne's notes on reductive groups http://jmilne.org/math/CourseNotes/RG.pdf p. 24. This implies that the algebraic groups are non-isomorphic, but not necessarily the statement you wanted.
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$\begingroup$ what is a root system and is there any simple answer? $\endgroup$– SinaCommented Jul 10, 2012 at 13:08
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4$\begingroup$ If SINA does not know root systems, perhaps the fastest argument is that $\textbf{SL}_2(\mathbb{F}_q)$ has a nontrivial center, while $\textbf{PGL}_2(\mathbb{F}_q)$ has trivial center. $\endgroup$ Commented Jul 10, 2012 at 13:09
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1$\begingroup$ @Jason Does $SL_2(2^k)$ has nontrivial center? $\endgroup$ Commented Jul 10, 2012 at 13:17
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1$\begingroup$ @Andrei -- I guess I was assuming that $q$ is odd. $\endgroup$ Commented Jul 10, 2012 at 14:16
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6$\begingroup$ Dear Timo, A couple of comments: (i) you probably mean root data rather than root systems; (ii) it's not clear (at least to me) that a statement about non-isomorphism of algebraic groups implies a corresponding statement about non-isomorphism of $\mathbb F_q$-points; (iii) related to point (ii), in fact the natural isogeny does induce an isomorphism if $q$ is a power of $2$ (this is easily checked, or see Jim Humphrey's answer), so point (ii) is a genuine concern, as far as I can tell. Regards, $\endgroup$– EmertonCommented Jul 10, 2012 at 14:44