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This may be an easy exercise but I am not getting it. Let $\mathbf F_q$ be a finite field with $q$ elements and $\mathbf F_{q^2}$ be its degree two extension. Define an automorphism $\sigma$ of $\mathbf F_{q^2}$ by $\sigma (x) = x^q$. For any matrix $A = (a_{ij}) \in M_n(\mathbf F_{q^2})$, let $A^{\star} = (a_{ji}^{\sigma})$ (i.e., $A^{\star} = A^{\sigma \mathrm{t}}$). Then finite unitary group $U_n(q)$ is a given by

$ U_n(q) = \left\lbrace A \in GL_n(\mathbf F_{q^2}) | A A^{*} = I_{n} \right\rbrace $

In paper of Wall (page 33), it is mentioned that order of this group is $q^{(n^2-n)/2} \prod\limits_{i=1}^{n} (q^i - (-1)^i)$. Question is how to prove this? Any help will be appreciated.

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There's a "classical" formula of Steinberg for order of $G(k)$ for any conn'd ss group $G$ over finite field $k$: see 11.16 in his "Endomorphisms of linear alg. gps". Steinberg imposes extra hypotheses on $G$, but unnecessary, as explained in 1.6 in Oesterle's paper "Nombres de Tamagawa" (which gives an elegant description of the formula). The inputs in formula are dimension of $G$, size of $k$, action of geometric Weyl group on geometric character gp of max. $k$-torus $T$, and Frob. action on this char. gp. There's a shorter cohomological proof. –  BCnrd Aug 3 '10 at 14:30
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A good reference for elementary proofs of such results is Wan Zhexian, Geometry of Classical Groups over Finite Fields, second ed., Science Press, Beijing/New York, 2002. –  Richard Stanley Aug 3 '10 at 14:56
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The question is really just a reference-request (for an old, standard result). The unified Lie-theoretic viewpoint is best, but there are nice older treatments such as Emil Artin's "The orders of the classical simple groups", Comm. Pure Appl. Math. 8 (1955). That was the year of Chevalley's famous Tohoku paper, to which Artin alludes. Calculating such orders of finite linear groups is not a trivial exercise, since it requires a strategy, but is by now fairly elementary. –  Jim Humphreys Aug 3 '10 at 15:08
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See Peter Cameron's notes on classical groups: maths.qmul.ac.uk/~pjc/class_gps . –  Robin Chapman Aug 3 '10 at 16:10
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Advice: These further comments on unified proofs of the most general order formula are not directly an answer about finite unitary groups, but eventually it's the direction to go. In the short run, if access is a problem, try one of the more classical sources mentioned already or if necessary online notes. Don't spend a lot of time rediscovering proofs of such old theorems. There are many things left to do in mathematics. –  Jim Humphreys Aug 3 '10 at 22:10
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1 Answer 1

This question from 2010 was just listed as "active", apparently because someone (not myself) downvoted it today. Anyway, rather than prolong the previous list of comments, I'll offer an explicit answer by pointing to an online resource that slightly predates Steinberg's 1968 AMS Memoir and is still a useful way to get into the details about Chevalley groups and their twisted analogues. In 1967-68, Steinberg gave a course at Yale which was written up by listeners and published in mimeographed form by the math department there. These typed notes are not as readable as typeset material and lack an index, but have been scanned in PDF format and placed on Steinberg's homepage at UCLA http://www.math.ucla.edu/~rst/

In particular, Section 11 deals with twisted groups, their Bruhat decompositions, and their orders. One advantage of this broader approach via Chevalley groups is that it makes transparent the close analogy between the order formulas for finite unitary groups and for special linear groups of similar size. In any case, as previous comments already indicate, the suggestion in the question that "This may be an easy exercise ..." is misguided. But the order formula itself is standard and well studied from a variety of directions in the literature cited.

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