From: mareg@primrose.csv.warwick.ac.uk () Subject: Re: The group SL_2(Z) Date: Fri, 13 Apr 2001 08:50:59 +0000 (UTC) Newsgroups: sci.math Summary: Commutator subgroup, automorphisms of SL_2(Z) In article <87pze8ak6pkb@forum.mathforum.com>, azmitamid@REMOVEmy-deja.com (Azmi Tamid) writes: >Does someone know if the group G= SL_2(Z) has non trivial Abelian >image i.e if the commutator group of G is the whole group , and also >what is the automorphism group of G ? > SL(2,Z) is generated by x = [ 0 1 ] and y = [ 0 1 ] [-1 0 ] [-1 -1 ] with defining relations x^4 = y^3 = 1, x^2 y = y x^2. (This is "well known". I can find you a reference if you like, but not right now because I am at home.) It follows that G/[G,G] is cyclic of order 12. The centre of the group has order 2, and is generated by x^2 = -I. The inner automorphism group is the group modulo its centre, which is PSL(2,Z), and is isomorphic to the free product C_2 * C_3. The full automorphism group is at least twice as big as that, because it contains PGL(2,Z), which is the image of integral matrices with determinant 1 or -1. I don't know off-hand whether PGL(2,Z) is the full automorphism group. Derek Holt. ============================================================================== From: Robin Chapman Subject: RE: The group SL_2(Z) Date: Fri, 13 Apr 2001 09:08:11 -0400 Newsgroups: sci.math >===== Original Message From mareg@primrose.csv.warwick.ac.uk () ===== [quote of previous message deleted --djr] > I don't know off-hand whether PGL(2,Z) is the full > automorphism group It isn't. By looking at the abelianization of G, we can see that the inner automorphism group has index at least 4 in the full automorphism group. Each automorphism of G induces one of G/[G,G], so we get a group homomorphism phi from Aut(G) to Aut(G/[G,G]). Each inner automorphism of G induces the trivial automorphism of G/[G,G], so the inner automorphisms of G lie in ker(phi). But G/[G,G] is cyclic of order 12, and so has 4 automorphisms. If we can realize all of these from automorphisms of G we are done, since then ker(phi) has index 4 in Aut(G). But that's easy to do from the presentation above. The 4 maps of the form x|-> x^{+-1} and y |-> y^{+-1} extend to automorphisms of G and they induce the four different autmomorphisms of G/[G,G]. Questions: 1. are all automorphisms of G trivial on G/[G,G] inner? 2. is there a nice way of describing what the above automorphisms of G do to arbitrary matrices in G? ------------------------------------------------------------ Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html "His mind has been corrupted by colours, sounds and shapes." The League of Gentlemen ============================================================================== From: mareg@primrose.csv.warwick.ac.uk () Subject: Re: The group SL_2(Z) Date: Sat, 14 Apr 2001 13:27:38 +0000 (UTC) Newsgroups: sci.math In article <3AF33FF5@MailAndNews.com>, Robin Chapman writes: [quote of previous message deleted --djr] >Questions: >1. are all automorphisms of G trivial on G/[G,G] inner? Yes I think they are. Here is a summary of a proof. Let H = G/Z(G) = C_2 * C_3 = * , say. It will be enough to prove that Aut(H) = Inn(H).2 (where the .2 comes from the automorphism a->a, b->b^-1), because then Aut(G) will map onto Aut(H) with kernel of order 2 (the automorphism x->x^-1, y->y of G). An automorphism of H maps a->a', b->b', where a', b' have order 2,3, and by the Kurosh subgroup theorem for free products, must be conjugate to and to in H. Hence, modulo the known subgroup Inn(H).2 of Aut(H), we can assume that a = a', and b' = b^g, for some g in H. But every element of H can be written in normal form (a) b^{e_1} a b^{e_2} ... a (b^{e_r}), where each {e_i} = 1 or -1, and the first and last terms may or may not occur. Since a and b' must generate H, and therefore b must be expressible in terms of a, b', we find that the only possible b' are b and aba, both of which give inner automorphisms. (For example if g = ab, b' = b^-1abab, then any element generated by a word in a and b' is already in normal form, and is clearly not equal to b.) >2. is there a nice way of describing what the above automorphisms >of G do to arbitrary matrices in G? I cannot think of any way to do this other than writing the matrix as a word in the generators - which is not particularly hard algorithmically, because it is more or less the same process as diagonalizing the matrix with row and column operations, but it is not what you would call a nice way of doing it. Derek Holt. ============================================================================== From: Robin Chapman Subject: RE: The group SL_2(Z) Date: Sun, 15 Apr 2001 05:08:23 -0400 Newsgroups: sci.math >===== Original Message From mareg@primrose.csv.warwick.ac.uk () ===== >In article <3AF33FF5@MailAndNews.com>, >> >>Questions: >>1. are all automorphisms of G trivial on G/[G,G] inner? > >Yes I think they are. Here is a summary of a proof. >Let H = G/Z(G) = C_2 * C_3 = * , say. > >It will be enough to prove that Aut(H) = Inn(H).2 >(where the .2 comes from the automorphism a->a, b->b^-1), >because then Aut(G) will map onto Aut(H) with kernel of order 2 >(the automorphism x->x^-1, y->y of G). > >An automorphism of H maps a->a', b->b', where a', b' have order 2,3, and >by the Kurosh subgroup theorem for free products, > must be conjugate to and to in H. This can also be seen by considering the action of PSL(2,Z) on the upper half-plane. >Hence, modulo the known subgroup Inn(H).2 of Aut(H), we can assume that >a = a', and b' = b^g, for some g in H. > >But every element of H can be written in normal form >(a) b^{e_1} a b^{e_2} ... a (b^{e_r}), >where each {e_i} = 1 or -1, and the first and last terms may or may not occur. > >Since a and b' must generate H, and therefore b must be expressible in >terms of a, b', we find that the only possible b' are b and aba, both >of which give inner automorphisms. OK. I belive this, but I'll have to work out the details. >(For example if g = ab, b' = b^-1abab, then any element generated by a >word in a and b' is already in normal form, and is clearly not equal to b.) >>2. is there a nice way of describing what the above automorphisms >>of G do to arbitrary matrices in G? > >I cannot think of any way to do this other than writing the matrix as a >word in the generators - which is not particularly hard algorithmically, >because it is more or less the same process as diagonalizing the >matrix with row and column operations, but it is not what you would call >a nice way of doing it. The basic "nontrivial" automorphism of SL(2,Z) has this form. Since its abelianization is cyclic of order 12, it has a unique homomorphism phi onto {1,-1}. Then A |--> phi(A)A is an automorphism of SL(2,Z). Now is there a simple description of phi .... ? ------------------------------------------------------------ Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html "His mind has been corrupted by colours, sounds and shapes." The League of Gentlemen