Showing Z(D_n) equals the intersection of centralizers C(x) and C(y)

[RJLiberator]

Homework Statement

If n ≥ 3, show that Z(D_n) = C(x) ∩ C(y).

Homework Equations

G is a group, g∈G
C(g) = {h∈G: hg = gh } The Centralizer of g
Z(G) = {h∈G: hg = gh for all g∈G} The center of G

∩ means the set of all points that fall in C(x) and C(y).

Every element of D_n can be uniquely written in the form y^i x^j.
x is a reflection and y is a rotation

The Attempt at a Solution

The elements of Dn are 1, y, y^2, y^2,...y^(n-1) and x, yx, y^2x, ..., y^(n-1)x
C(x) = { h∈G: hx = xh }
C(y) = { h∈G: hy = yh }
Z(D_n) = {h∈G: hx=xh for all x∈G}

Note: xy^i = y^(n-i)x for 1 ≤ i ≤ n-1

Okay. I need some help.

I don't understand how to use the definitions of C(x) and C(y) and Z(D_n) here.

 

[andrewkirk]

To prove equality, you need to prove inclusion in both directions:
1. Prove that $Z(D_n)\subseteq C(x)\cap C(y)$
2. Prove that $C(x)\cap C(y)\subseteq Z(D_n)$

The first one is easier, so I suggest you do that first. Pick an element $a\in Z(D_n)$. Is it in $C(x)$ (look at the definitions)?. Then see if it's in $C(y)$.

 

[RJLiberator]

My problem is, I don't understand the definitions...

C(g) = {h∈G: hg = gh } The Centralizer of g
Z(G) = {h∈G: hg = gh for all g∈G} The center of G

If we have a existing in Z(Dn) then a is clearly abelian, but then we need to show that it either exists in C(x) or C(y).
C(y) says hy=yh and C(x) says hx=xh
If a = xg=gx then we just need to prove that g exists in G?

 

[andrewkirk]

C(g) is the set of all elements of G that commute with element g.
Z(G) is the set of all elements of G that commute with everything.
Remember that x and y were given specific meanings in the above definition of the dihedral group. They are the names of specific group elements, not just variable names.