# What is the subgroup generated by involutions?

I was recently taking some notes on the Cartan-Dieudonné theorem: if $(V,q)$ is a nondegenerate quadratic space of finite dimension $n$ over a field of characteristic not $2$, then every element of the orthogonal group $O(q)$ is a product of at most $n$ reflections through nondegenerate hyperplanes. (Moreover -- this is not the hard part! -- $n$ is sharp, as one sees by considering $-1$.)

So that's how things go in $O(q)$. At some point my mind wandered and I started thinking about $\operatorname{GL}(V)$ instead. Hyperplane reflections don't seem to have the same significance here, so I started to think about involutions -- i.e., elements of order $2$ -- instead. What is the subgroup of $\operatorname{GL}(V)$ generated by all the involutions?

Well, I was smart enough to realize that every involution has determinant $\pm 1$, so the subgroup in question is definitely contained in the subgroup of all matrices with determinant $\pm 1$ -- let's call this $\operatorname{GL}(V)^{(2)}$ -- hence usually proper. But is it actually this whole group? I checked one example: $\operatorname{GL}_2(\mathbb{F}_3)$ is isomorphic to $S_4$ [not even close -- but still, one can see that my conclusion is true!], which is indeed generated by its involutions.

I didn't make much more progress than that, so I started googling. Eventually I came across the following paper:

Gustafson, W. H.; Halmos, P. R.; Radjavi, H. Products of involutions. Collection of articles dedicated to Olga Taussky Todd. Linear Algebra and Appl. 13 (1976), no. 1/2, 157–162.

Apparently in this paper the authors prove, among other things, that for any finite-dimensional vector space over any field, any element of $\operatorname{GL}(V)^{(2)}$ is a product of at most $4$ involutions. This is a pretty striking result, so I have some questions.

Question 1: How do you prove it? (I can't immediately access the paper, and unfortunately the description in the MathSciNet review was not immediately illuminating to me.)

Question 2: Can this result really not have been known until 1976?? The Cartan-Dieudonné theorem was proved at the latest in $1945$, the first publication date given for Dieudonné's book on classical groups. No one wondered about general linear groups for another $30$ years??

Question 3: In lots of generality we can take a group $G$ and ask what is the subgroup generated by its involutions. To fix ideas, suppose that $G$ is the group of $K$-rational points of a connected linear algebraic group over a field $K$. (Or choose whatever special case of this you like.) Can we say something about the subgroup generated by involutions? What about the least number $N$ of involutions so that every element of this subgroup is a product of at most $N$ involutions (if such an $N$ exists)?

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Note that the subgroup generated by all involutions is normal which is a very strong condition for groups with few normal subgroups (such as say semi-simple algebraic groups). –  Torsten Ekedahl Nov 28 '10 at 16:25
Pete, ${\rm{GL}}(V)^{(2)}$ is not much more general than ${\rm{SL}}(V)$ (i.e., suffices to handle ${\rm{SL}}(V)$). If $G$ is a split connd ss group over a field $k$ then the structure of the open Bruhat cell and Weyl group and Bruhat decomposition (as old as the hills, so to speak) gives that $G(k)$ is generated by some ${\rm{SL}}_2(k)$'s and ${\rm{PGL}}_2(k)$'s, one per pos. simple root (all ${\rm{SL}}_2(k)$'s when $G$ is simply connd). So for #3 it suffices to treat ${\rm{SL}}_2(k)$ and ${\rm{PGL}}_2(k)$, getting a crude bound on $N$. Try the Book of Involutions for good bounds? –  BCnrd Nov 28 '10 at 16:36
SL(2,k) is not generated by its involutions if k has characteristic ≠ 2. SL(2,q) for q odd has quaternion Sylow 2-subgroups, so only a single involution. –  Jack Schmidt Nov 28 '10 at 17:12
Dear Jack: that was dumb of me (alt. proof: non-scalar involution in GL$_2(k)$ away from char. 2 is semisimple and hence conjugate to diag(1,-1), so not in SL$_2(k)$). And of course there are problems for $G$ whose fundamental group is not 2-torsion, since for finite $k$ by Lang's theorem $G(k)$ maps onto H$^1(k,\mu_G)$ with $\mu_G$ the Cartier dual to the fundamental group. For instance, PGL$_3(k)$ maps onto $k^{\times}/(k^{\times})^3$ killing all involutions. So let's focus on simply connected $G$. Then on to rank 3: SL$_3$, Sp$_4$, and G$_2$. Takers? –  BCnrd Nov 28 '10 at 17:32
@Torsten: that's an absolutely key point, thanks. This makes me wonder: can one show that the subgroup of $G$ (the $K$-rational points of a connected linear group) generated by the involutions is necessarily Zariski-closed? –  Pete L. Clark Nov 28 '10 at 19:00

Maybe I can comment on Question 2. To me the essential point is that this kind of result belongs to elementary linear algebra and basic group theory rather than to geometric algebra. Generation of special linear groups (or slightly larger groups) by involutions probably goes back a long way and may have multiple origins, though I don't have a definitive reference handy. (I'd probably try to ask someone like Gustafson or Djokovic rather than search the scattered literature.)

From a semi-modern viewpoint, for instance, generation of a general linear group over an arbitrary field focuses on the traditional building blocks: elementary, permutation, and diagonal matrices. Since a one-parameter group of elementary matrices along with a suitable permutation matrix will generate a copy of SL$_2$, generation of a special linear group just requires these two kinds of ingredients. Here a permutation matrix is obviously a product of involutions. On the other hand, SL$_2$ is almost simple, while its subgroup generated by involutions is obviously normal. (To include matrices of determinant -1 is a further step.)

Books like those by Dieudonne (and Artin) mix in further ideas arising from the geometry of a bilinear form, along with consideration of exact upper bounds on the number of involutions needed to generate any group element. Such upper bounds for products of involutions in general linear groups may have come along later.

P.S. I used SL$_2$ subgroups in the rough sketch above just to emphasize some elementary steps for showing that the big groups in question are generated by involutions; of course, simplicity of SL$_n$ modulo its center for most fields could be used directly here, plus ad hoc arguments for the few nonsimple cases. Then multiplying any matrix of det -1 by a diagonal involution reduces to the det 1 case. Anyway, none of this older theory deals with an explicit upper bound on number of involutions required.

FURTHER UPDATE: Sorry to have confused matters with my offhand remarks here. While my "essential point" is unchanged, a quick look at the paper quoted (along with other comments here) indicates that the statement of the theorem is true: Each matrix of determinant 1 or -1 is the product of at most four involutions (exactly four if you follow their convention that an involution is an element of square 1). I managed to make a copy of the bound journal article at the library this morning, where the methods used rely just on linear algebra. The authors are following in a tradition where products of 2 or 3 matrix involutions have been characterized.

The point seems to be that you work exclusively in the larger group of all matrices having determinant 1 or -1 (leaving special linear groups aside). Here the argument involves rational canonical forms and permutation matrices having either determinant. With this flexibility, the argument in the paper seems OK but is written down a bit loosely. It would help to state their theorem by giving the group in question a name (there seems not to be a standard one). Since I think first in terms of connected algebraic or Lie groups, I was too quick to interpret the theorem as applying to special linear groups. (Like Brian I am also tempted to consider other Lie types, relying on Bruhat decomposition and the like.)

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I'm confused: as in the above comments, isn't $-1$ the only involution in $\operatorname{SL}_2$? –  Pete L. Clark Nov 30 '10 at 0:09
Dear Pete: sure, but one can get at all split simply connected examples other than ${\rm{SL}}_2$ by using the rank-3 cases (SL$_3$, G$_2$, and Sp$_4$), so have you looked into those basic building blocks instead? –  BCnrd Nov 30 '10 at 4:08
@BCnrd: not yet, no. My point is that I find the sentence "On the other hand, $SL_2$ is almost simple, while its subgroup generated by involutions is obviously normal and too big to be the center" to be confusing. (Or, less politely but more plainly, it seems not to be true.) –  Pete L. Clark Nov 30 '10 at 13:11
Dear Pete: Yes, it was a small goof. But just move on: when rank-2 is too small, a standard device is to use the rank-3 cases to handle everything else. That is why (in my various comments) I keep suggesting to look into those basic rank-3 cases, which would then handle everything else split and simply connected by using Bruhat decomposition, torus centralizers, and the like (as Jim says). –  BCnrd Dec 1 '10 at 9:36

You might want to check out:

D.Z. Djokovic and J. Malzan , Products of reflections in the general linear group over a division ring. Linear Algebra Appl. 28 (1979), pp. 53–62.

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@Igor: Thanks. It looks like this paper deals with the case of reflections (i.e., involutions with exactly one negative eigenvalue) in $GL(V)$ rather than involutions. Here, as in the Cartan-Dieudonne Theorem, the minimal number seems to depend on the dimension of $V$. This is not exactly what I asked about, but it is just as interesting and makes me wonder even more about Question 2: apparently Dieudonne's book on "classical groups" gives short shrift to $GL(V)$! –  Pete L. Clark Nov 28 '10 at 19:08

I am unable to access this paper, but it also seems relevant:

Knüppel, F. Products of involutions in orthogonal groups. Annals of Discrete Mathematics 37 (1988), pgs. 231-247.

From the abstract: "Let $G$ be a group and $S$ a set of involutions generating $G$. (a) What is the minimal number $k$ such that every element of $G$ is a product of at most $k$ elements of $S$? (b) Given some $\pi \in G$, what is the minimal number $k$ such that $\pi$ is a product of $k$ elements of S? We discuss both questions for particular groups $G$ and subsets $S$."

Edit: Actually, minus some pages, most of the paper is available on Google Books.

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