Question

Let n>=2, p a large prime, G = SL_n(Z/pZ).

If n=2, there are words that, while not conjugate in the free group, do have identical trace in G. For example, tr(g h^2 g^2 h)= tr(g^2 h^2 g h) for all g, h in SL_2(Z/pZ).

Question: does the same happen for n>=3? Could it even be possible that there are words w_1, w_2 that are not conjugate even in G=SL_n(Z/pZ), yet always have the same trace: tr(w_1(g_1,g_2,..,g_k)) = tr(w_2(g_1,g_2,...,g_k)) for all g_1,...,g_k in SL_n(Z/pZ)?

This would be extremely helpful.

Actually, I just need a weaker statement:

Wild guess.- Let g, h be elements of SL_n(K), n>2. There is a constant k (which may depend on n but not on K) such that there are two elements a, b of the ball ({g,h,g^{-1},h^{-1},e})^k for which (1) tr(a)=tr(b) and (2) a is not conjugate to b (for g and h generic).

Does anybody have a clue as to whether this is or isn't true?

Answer 1

I don't think Critch's reply above answers Harald's question; it seems to presume that the map F_2 -> SL_n(Z/pZ) factors through a chosen inclusion of SL_2(Z/pZ), while Harald wants pairs of elements in F_2 (or, more generally, F_k) which have the same trace after applying ANY homomorphism F_k > SL_n(Z/pZ).

EDITED to comment on David's answer below:

"Also, if there is any finite quotient G of F_k, in which the two elements stay nonconjugate, then there is some representation of G in which they have different traces."

Yeah, but not an n-dimensional representation.

Answer 2

Since James seems to have found the group theory lemma we need, I'm going to rewrite this post to be clearer about what it proves.

Theorem: If g and h are two non-conjugate elements of the free group F_k, then there is some n, some p, and a map r:F_k --> GL_n(Z/p) such that Tr r(g) \neq Tr r(h).

Proof: By James's post, there is a finite group G and a map a: F_k --> G so that a(g) and a(h) are not conjugate. As characters distinguish conjugacy classes, there is a cyclotomic field K=Q(e^{2 pi i/q}) and a representation b: G --> GL_n(K) such that Tr b(a(g)) \neq Tr b(a(h)). Let p be a prime which is (1) large enough not to appear in the denominators of any of the matrices occurring in b(G) (2) congruent to 1 modulo q and (3) large enough not to divide Tr b(a(g)) - Tr b(a(h)). By condition (2), there is a prime P of K, living over p, such that O(K)/P = Z/p. By condition (1), we can reduce the entries of the matrices of b(G) modulo p. This gives us a map c : G --> GL_n(Z/p) which is morally the composite of G --> GL_n(K) --> GL_n(O(K)/P). (The second map of this sequence does not actually exist.)

Composing a and c gives the map r. Condition (3) insures that Tr r(g) \neq Tr r(h). QED

Note that I have not solved the following more difficult question: given n, do there exist any pair of nonconjugate elements g and h in F_k so that Tr r(g) = Tr r(h) for all maps r: F_k --> GL_n(Z/p).

Answer 3

I'm not sure I follow the reduction to the assertion about non-conjugate elements in free groups. However, it is true that, if two members of a free group are not conjugate, then there is a finite quotient in which their images are not conjugate. This property is known as conjugacy separability, and it is well known that free groups are conjugacy separable. There are a number of ways to prove this. One can, for example, reduce to finitely generated nilpotent groups via the lower central series in the free group (which is residually a torsion free nilpotent groups) and then invoke the result of Formanek/Remeslennikov that f.g. torsion-free nilpotent groups are conjugacy separable. Alternatively, free products of conjugacy separable groups are conjugacy separable (Remeslennikov, 1971), and infinite cyclic groups obviously are.

Answer 4

Yes, if I understand your question correctly. Consider SL_2(Z/pZ) included in SL_n(Z/pZ) as those elements with a 2x2 block in the top left, 1's on the the remaining diagonal, and 0's elsewhere. Also consider The free group F_2 on g_1, g_2 included into the free group F_n on g_1, g_2, ..., g_n.

Two elements of F_2 are conjugate in F_n iff they are conjugate in F_2 (since if you conjugate by a word involving some g_i, i>2, then you land outside F_2). So whatever pairs of elements work in your n=2 example also work for n arbitrary: they will have the same trace in SL_n(Z/pZ), namely [their trace in SL_2(Z/pZ)] + [n-2], the extra diagonal elements we added, but the words expressing them won't be conjugate in F_n because they weren't in F_2.