The conjugacy classes which appear in a subgroup

A fairly well-known fact about finite groups says that if H is a subgroup of G, and H intersects every conjugacy class in G, then in fact H=G. This is quite useful, for instance, for some problems of Galois theory, because one might have to understand a finite group using information only about which conjugacy classes it represents in a bigger group (e.g., a Galois group represented as permutation groups of the roots of an integral polynomial, where the factorization of the polynomial modulo various primes indicates which conjugacy classes of the corresponding symmetric group intersect the Galois group; I’ve already mentioned this type of things here and here).

It is natural to ask what happens with other kinds of groups. The example of compact Lie groups shows that if G is infinite, there may well exist a subgroup H intersecting every conjugacy class; for instance, if G=U(n), every element can be diagonalized, i.e., every element is conjugate to one in the subgroup H of diagonal matrices (which, if n is not 1, is not the same as G…) However, these are quite special groups, and one might suspect that some interesting infinite groups retain this property (which I’ll call the Jordan property here, as suggested by Serre’s nice paper about this theorem of Jordan).

Although I’ve started looking around, I haven’t found much information yet on this. The first groups I’m trying to understand are arithmetic groups like G=SL(n,Z). Here’s one simple example in such a case: if n is at least 3, then G has the Jordan property “with respect to finite index subgroups” (i.e., any finite index subgroup intersecting all conjugacy classes of G is equal to G). This requires a fairly big hammer, but is otherwise very easy: by the Congruence Subgroup Property, any H of finite index satisfies

G(d)\subset H\subset G,

for some integer d, where

G(d)=\{g\in G\,\mid\, g\equiv 1\text{ mod }d\}

is a principal congruence subgroup. This means that, for some subgroup Γ of the finite quotient G/G(d), we have

H=\{g\in G\,\mid\,  g\text{ mod } d \in \Gamma\},

but then it is immediate (by lifting to H) that if H intersects all conjugacy classes of G, then also Γ intersects every conjugacy class in G/G(d), and we get


from the finite group case, and therefore H=G.

More generally, one sees at least (without using the Congruence Subgroup Property) that if H is a subgroup of G=SL(n,Z) intersecting every conjugacy class, then we have

H\text{ mod } d=SL(n,\mathbf{Z}/d\mathbf{Z}),

for all d (because the reduction modulo d maps G surjectively to SL(n,Z/dZ) for all d). However, this condition is not as stringent as it may look: it is known (the “Strong Approximation Theorem” of Mathews, Vaserstein and Weisfeiler) that this holds, at least for all integers d coprime with some “conductor” f, for any subgroup of SL(n,Z) which is Zariski-dense in SL(n), and such groups can be quite “small”. However, one might intuitively hope that, being “smaller” than finite index subgroups, they would intersect fewer conjugacy classes (?). On the other hand, I also don’t know offhand of a non-trivial subgroup with conductor f=1

For the special case n=2, the Congruence Subgroup Property fails (one way to see it, as explained in this survey of Raghunathan, is to contrast the fact that SL(2,Z) has finite quotients like the alternating group A5, whereas any non-abelian simple quotient of a congruence subgroup is of the type SL(2,Z/pZ) for a prime p, and none of these is isomorphic to A5, simply because none is of order 60). Then it’s not clear to me if some finite index subgroup (not of congruence type) could intersect every conjugacy class of SL(2,Z).

Hopefully, I’ll have the occasion to write more about this as I explore the literature…

Published by


I am a professor of mathematics at ETH Zürich since 2008.

5 thoughts on “The conjugacy classes which appear in a subgroup”

  1. I claim that for any group G, if H is a finite index subgroup which meets every conjugacy class, then H=G.

    Proof: Let [G:H] = n and let K be the kernel of the map G –> S_n given by the action on G/H. So K is normal in G, G/K is finite and K \subseteq H. My point is that these are the only properties of congruence subgroups which you used.

    Specifically, H/K meets every conjugacy class in G/K. So, by the theorem for finite groups, H/K = G/K. Since H contains K, this implies G=H.

    Am I missing something?

  2. You’re right (I now seem to remember that this fact was mentioned in Serre’s paper)!

    In the more general (infinite index) case, one gets in fact that the subgroup surjects to every finite quotient. Something must be known about groups where such a thing is true…

  3. The proof is fairly simple, though a bit clever; first, the number of distinct conjugates xHx-1 of the subgroup H is at most its index in G; then the assumption is that the union of those conjugates is the whole group G; since each has |H| elements, this would only be possible if the conjugates are disjoint sets, but since each of them contains the identity element, this is in turn only possible if there a unique conjugate, i.e., if H=G.

    Also, a correction to my previous comment: the place where I had seen the statement for arbitrary finite-index subgroups was not in Serre’s paper, but in his (unpublished) lecture notes on analytic number theory.

  4. Minor quibble: A_5 = PSL_2(F_5) _does_ arise as a congruence quotient of PSL_2(Z) (and SL_2(Z)). I had undergraduates (as a summer project) prove that A_n is a quotient of PSL_2 = if and only if n = 1,2,3,4,5 or n is at least 9. The latter are certainly non-congruence quotients.

Leave a Reply

Your email address will not be published. Required fields are marked *