Thread: 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.

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.

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.

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).