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^{-1}y is in Z(g). so x^{-1}y commutes with g:
x^{-1}yg = gx^{-1}y
x^{-1}ygy^{-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^{-1}y commutes with every element g of G, it surely lies in Z(G).