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This question is inspired by the recent question "The finite subgroups of SL(2,C)". While reading the answers there I remembered reading once that identifying the finite subgroups of SU(3) is still an open problem. I have tried to check this and it seems it was at least still open in the Eighties.

Can anyone confirm or deny that the finite subgroups of SU(3) are not all known? And if this is true, then what is the source of the difficulty?

Secondly, what is known of the finite subgroups of SU(n) for n > 3?

UPDATE: Thanks to those below who have corrected my ignorance! It seems that I may have been tricked by some particularly sensationalised abstracts (or perhaps just misunderstood them.)

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This discussion has been cited in the review by Shengkui Ye of Bela Bauer and Claire Levaillant, A new set of generators and a physical interpretation for the SU(3) finite subgroup D(9,1,1;2,1,1), Quantum Inf. Process. 12 (2013), no. 7, 2509–2521, MR3065503. –  Gerry Myerson Sep 15 '14 at 3:50

7 Answers 7

up vote 30 down vote accepted

There is an algorithm due to Zassenhaus which, in principle, lists all conjugacy classes of finite subgroups of compact Lie groups. I believe that the algorithm was used for $\mathrm{SO}(n)$ for at least $n=6$ if not higher. I believe it is expensive to run, which means that in practice it is only useful for low dimension.


Now that I'm in my office I have my orbifold folder with me and I can list some relevant links:

  1. Zassenhaus's original paper (in German) Über einen Algorithmus zur Bestimmung der Raumgruppen
  2. There is a book by RLE Schwarzenberger N-dimensional crystallography with lots of references
  3. There are a couple of papers in Acta Cryst. by Neubüser, Wondratschek and Bülow titled On crystallography in higher dimensions
  4. There is a sequence of papers in Math. Comp. by Plesken and Pohst titled On maximal finite irreducible subgroups of GL(n,Z) which I remember were relevant.

Independent of this algorithm, there is some work on $\mathrm{SU}(n)$ from the physics community motivated by elementary particle physics and more modern considerations of the use of orbifolds in the gauge/gravity correspondence.

The case of $\mathrm{SU}(3)$ was done in the mid 1960s and is contained in the paper Finite and Disconnected Subgroups of SU(3) and their Application to the Elementary-Particle Spectrum by Fairbairn, Fulton and Klink. For the case of $\mathrm{SU}(4)$ there is a more recent paper A Monograph on the Classification of the Discrete Subgroups of SU(4) by Hanany and He, and references therein.

Further edit

The paper Non-abelian finite gauge theories by Hanany and He have the correct list of finite subgroups of SU(3), based on Yau and Yu's paper Gorenstein quotient singularities in dimension three.

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The Fairbairn, Fulton and Klink paper has at least one error: they claim that $A_6$ appears as a finite subgroup of $\mathrm{SU}(3)$, but $A_6$ has no 3-dimensional representations, as indeed their character table shows. What is the group of order 360 that appears? –  Dylan Thurston Apr 1 '10 at 6:24
In fact, this statement already appears in the book of Blichfeldt, Dickson and Miller which Bruce Westbury mentions in his answer. It is claimed that this group is simple. There is a thread in sci.math.research claiming that any simple group of order 360 is isomorphic to $A_6$. If so, then the mistake is in claiming that the group is simple, because as you point out, $A_6$ has no nontrivial representation of dimension 3. (The smallest is of dimension 5.) I must admit that I have not gone through their proof, so cannot say where the error lies. –  José Figueroa-O'Farrill Apr 1 '10 at 22:32
I forgot to link to the sci.math.research thread. Here it is: –  José Figueroa-O'Farrill Apr 2 '10 at 0:44
Apparently it is sci.math, not sci.math.research -- sorry. How I wish I could edit comments! –  José Figueroa-O'Farrill Apr 2 '10 at 0:44
Reading a little more closely, it seems that the group with the 3-dimensional representation is not $A_6$, but rather a $\mathbb{Z}/3$ central extension of $A_6$. –  Dylan Thurston Apr 2 '10 at 6:07

The finite subgroups of SU(3) have been known for a century. I think you can find it in these references (my Departmental library does not go back this far):

MR1500676 Blichfeldt, H. F. On the order of linear homogeneous groups. II. Trans. Amer. Math. Soc. 5 (1904), no. 3, 310--325.

MR1511301 Blichfeldt, H. F. The finite, discontinuous primitive groups of collineations in four variables. Math. Ann. 60 (1905), no. 2, 204--231.

MR1560123 Blichfeldt, H. F. Blichfeldt's finite collineation groups. Bull. Amer. Math. Soc. 24 (1918), no. 10, 484--487.

and also in this book

MR0123600 (23 #A925) Miller, G. A. ; Blichfeldt, H. F. ; Dickson, L. E. Theory and applications of finite groups. Dover Publications, Inc., New York 1961 xvii+390 pp.

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The link to the first paper of Blichfeldt's in your list is: –  José Figueroa-O'Farrill Mar 4 '10 at 13:38
The book can be downloaded off of Google Books at:… –  Steve D Mar 4 '10 at 14:34
Really? I can't seem to: is this dependent on geography? I'm in the UK. –  José Figueroa-O'Farrill Mar 4 '10 at 15:24
Yes, I'm in the U.S. and can download it. I've seen the same thing happen before with a different book to someone in Australia, so I know it can be geography dependent. It seems Google has verified that it is in the public domain in the U.S., but either it isn't public domain in the UK or Google doesn't know and they're just being careful. –  Jonas Meyer Mar 5 '10 at 3:44
José, try this:… –  Jonas Meyer Mar 5 '10 at 3:48

Bit late with this but since I didn't see it above I will mention it: The paper by W. Feit MR0427449 (1970) says that finite linear groups up to dimension $7$ have been classified.

He has a list of primitive subgroups of $G\subset SL(n,\mathbb{C})$ with $Z(G)\subset G^\prime$ up to $n=7$. Probably this has been improved upon.

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You can laso have a look at for the case of $SL_3$. It discusses an algebraic form of the McKay correspondence in dimension 3. –  DamienC Feb 4 '12 at 10:07

This addresses the second question "What is known about finite subgroups of $SU(n)$". A special case of the Margulis lemma implies that for each $n$, there is an $m(n)$ such that any finite subgroup of $O(n)$ has an abelian subgroup of index $m(n)$ (see Corollary 4.2.4 of Thurston's book). Thus, there is a normal abelian subgroup of index at most $m!$. So one may make a statement: there are finitely many finite groups so that any finite subgroup of $SU(n)$ is an abelian extension (of rank at most $n-1$) of one of these finitely many groups. It would be quite interesting to obtain an estimate of the function $m(n)$, which should be possible by giving an effective proof of Margulis' theorem. I did a literature search once to see if anyone had attempted this, but I didn't find anything, and I would be curious if someone knows something.

Addendum: Working backwards from Weisfeiler's paper referenced in Keivan's comment, I found a result of Collins implies that a finite linear subgroup of $GL(n,C)$ has an abelian normal subgroup of index at most $(n+1)!$ when $n\geq 71$ (and gives the bound for all $n$). Since finite subgroups of $GL(n,C)$ are conjugate into $U(n)$, this bound works for $SU(n)$. See also Collins paper on primitive representations, which has some historical discussion of this problem.

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It is possible to get different estimates with different methods. I think the best one so far (which uses Classification) is due to Weisfeiler. Here is a link to the article: –  Keivan Karai Jun 7 '10 at 17:05
Thanks for the reference, I didn't know this was called Jordan's theorem. I suppose the estimate of $m(n)$ could be much smaller, however (where one does not require a normal subgroup). –  Ian Agol Jun 7 '10 at 17:26

This is really a comment on the top answer above, but since new users can't comment, I'll let someone else manually transfer the information to the right place.

There is a further mistake in the list of Fairbairn, Fulton and Klink (repeated in the list of Hanahy and He), which appears to be a misunderstanding of the classification by Blichfeldt et al. Two of the cases in that classification consists of semidirect products of abelian groups by $A_3$ and $S_3$. However, it is not specified which abelian groups can occur in this fashion!

Fairbairn, Fulton and Klink mistakenly assume that the abelian group in question has to be $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n \mathbb{Z}$ for some positive integer $n$, thus giving rise to the groups they denote $\Delta(3n^2)$ and $\Delta(6n^2)$. However, this is not the case.

Example 1: $A_3$ acts on the copy of $\mathbb{Z}/7\mathbb{Z}$ generated by the diagonal matrix with entries $e^{2\pi i/7}, e^{4 \pi i/7}, e^{8 \pi i/7}$; this example occurs inside the exceptional subgroup of order 168. More generally, if $m,n$ are positive integers and $m^2+m+1 \equiv 0 \pmod{n}$, then $A_3$ acts on the copy of $\mathbb{Z}/n\mathbb{Z}$ generated by the diagonal matrix with entries $e^{2\pi i/n}, e^{2m \pi i/n}, e^{2m^2 \pi i/n}$.

Example 2: $S_3$ acts on the copy of $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z}$ generated by the diagonal matrices with entries $e^{2\pi i/9}, e^{2\pi i/9}, e^{14 \pi i/9}$ and $1, e^{2\pi i/3}, e^{4\pi i/3}$; this example occurs inside the exceptional subgroup of order 648.

I don't know a reference for the complete classification of the abelian groups that can occur inside the semidirect product. Yau and Yu don't say any more than Blichfeldt et al, though they do at least provide a helpful rewrite of the classification in modern language.

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Check out

He has used GAP to list out the groups with a faithful 3D irreducible representations....That way he misses out on some subgroups of SU(2) well as abelian groups...which might have a faithful 3D representation which is reducible....

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Early on in the paper, the author states that he restricts his attention to those finite subgroups of U(3) which act irreducibly on the fundamental representation, hence it is somewhat misleading to say that "he misses out on...". –  José Figueroa-O'Farrill Sep 7 '10 at 12:55
Yes..I am sorry for the loose statement..thanks for clarifying the point... –  krishnamohan Sep 30 '10 at 7:50

Please see

We have classified all groups of order less than 100 into subgroups of U(2), SU(2), U(2)XU(1), SU(2)xU(1), U(3) and SU(3)...

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