# Cosets

• Oct 24th 2012, 09:27 PM
TheHowlingLung
Cosets
1. For each g in G, show that y is an element of xZ(g) (left coset of the centralizer) if and only if ygy-1 = xgx-1

2. Show that y is an element of xZ(G) (left coset of the center) if and only if ygy-1 = xgx-1 for all g in G.
• Oct 24th 2012, 09:46 PM
johnsomeone
Re: Cosets
1) Do you know the definition of the centralizer of g?
2) Do you know the definition of the center of a group?
3) Do you know the definition of a coset?
These two problems are just straightforward consequences of the definitions.
Maybe this will help: a is in bH (where H is a subgroup of some group G, a and b are elements of G) if and only if (b-inverse)(a) is in H.
• Oct 25th 2012, 03:47 AM
TheHowlingLung
Re: Cosets
I know the definitions, I'm just having trouble seeing how to use them to show that those statements are true.

The centralizer of g in G is the set of elements of G which commute with g. The center of a group is the set of elements that commute with every element of G. A coset is basically a shifted subgroup of G, more specifically gH = {gh: h E G} where H is a subgroup of G.

I know how to find centralizers, centers, and cosets of a group, but I'm not really sure how to show these proofs.
• Oct 25th 2012, 08:12 AM
Deveno
Re: Cosets
this can be rephrased as follows:

different conjugates of g give rise to distinct cosets of Z(g). think about what this may mean. anyway:

suppose y is in xZ(g). what does this mean? it means x-1y is in Z(g). so x-1y commutes with g:

x-1yg = gx-1y
x-1ygy-1 = gx-1 (multiplying on the right by y-1)
ygy-1 = xgx-1 (multiplying on the left by x).

that's what we wanted to prove for y in xZ(g) implies ygy-1 = xgx-1.

the other implication merely runs this argument backwards.

that's part (1) of your problem.

part (2) is an easy generalization. for if x-1y commutes with every element g of G, it surely lies in Z(G).