# SUBSET K of elements in a group with finite distinct distinct conjugates

1. Nov 30, 2011

### Bachelier

WTS, is that such set is a subgroup.

I need to show closure under group operation and inverse.

I can do the inverse which is usually the hardest part, but I'm stuck on the grp op.

So let a in K and b in K, both have finite distinct conjugates. Their conjugates are in the group too. WTS is that ab in K too.

if a, b in K then xax-1 = c in K and yay-1

consider xy(ab)(xy)-1 note since xax-1 and yay-1 are finite and distinct then x and y are finite and distinct hence xy and its inverse is finite

hence xy(ab)(xy)-1 in K

what do you think?

2. Dec 1, 2011

### Deveno

it looks like you're confusing elements and sets.

the SET of all conjugates of a {xax-1: x in G} is finite, not the conjugates themselves.

so you want to show that if the set of conjugates of a is finite, and the set of conjugates of b is finite, so is the set of conjugates of ab.

it might be helpful to note that if x is ANY element of G:

x(ab)x-1 = (xax-1)(xbx-1).