Recent content by Ultraworld

well, if you can prove all the 3 generators are of finite order than I can finish this question. Because Suzuki has this exercise in his book where he assumes G is finite and he also gives lots of hints. Here however I can not a priori assume G is finite.

I doubt wetter ToddCoxeter CE will help me :frown:

This is an exercise from a book from Serre called Trees. Given the group G = < a, b, c | bab−1 = a2, cbc−1 = b2, aca−1 = c2 > I have to prove G = 1. I don't have a clue. Of course G' = G (commutator subgroup equals the group itself) but I don't know what to deduce from that. Another...

yes but now in my exercise I have that the elements of the image are of the form cI for a scalar c in C. Exactly what I want cause now I got my contradiction.

I got a result in my book which says I am partly right. Given an irreducible representation X : G -> GL(n, C). Suppose that a square matrix M commutes with X(g) for every g in G. Then M = cI with c a scalar. I can use this very well in my homework. thanks for your help anyway Matt. Appreciated

:frown: an easy counterexample is indeed the group generated by the square matrix 1 0 1 -1

Big appologies but the abelian subgroup has to be finite. In that case I do really think it is true

hmm, thanks for the info Matt. I thought I tackled the original question but I have to look at it again I am afraid.

wait a minute, one of my algebra books mention this as a theorem I think. I ll have a look at it. edit: well, they only say something about the center.

How do finite abelian subgroups of GL(n, C) with n > 1 look like ? I would say the elements of those subgroups are only the diagonal matrixes but I am not sure (for my homework I do not have to prove it but I want to use this result if it is true). GL(n, C) are all the invertible matrixes over...

proof of proposition: Given arbitrary g in G. Given x and y in G such that xC(g) != yC(g). So C(g) != x-1yC(g) so x-1y not in C(g). Now assume xgx-1 = ygy-1. We easily derive from this g = (x-1y)g(x-1y)-1. So x-1y in C(g). Contradiction so xgx-1 != ygy-1. so the order of...

I suspect the following proposition holds proposition: Given arbitrary g in G. Then the order of the conjugacy class which contains g equals [G : C(g)]. edit: http://en.wikipedia.org/wiki/Conjugacy_class says I am right. :-) Now I am happy cause I do have some structure for that...

Thanks for the help guys. Now try wetter I can do this for a non-abelian group of order 48. Because that is the real exercise. This was just a example in the book (but they did not explained why they got this all)

ok here we go Take arbitrary g in G. We define the subgroup C(g) = {x in G : xg = gx}. We got Z(G) < C(g) < G. So |C(g)| = 3, 9, 27. Now assume g not in Z(G). We make case distinction order 3 is impossible cause g in C(g) but g was not in Z(G). order 9, we leave this one empty. order 27...

I still do not know why there are 11 conj-classes. :frown: Im afraid I just do not know enough small tiny theorems to prove it. I only know The identity element is in a conj-classs of its own. The order of a conj-class devides the group order. So if I could prove there are no...