Centers and Centralizers

The Center of a group Definition: Let G be a group. The center of G, denoted Z(G) is the set of group elements that commutes with every element of G. That is,

Z(D4) • R0 R90 R180 R270 H V D D'

Z(D4) contains R0 • R0 R90 R180 R270 H V D D' R0 commutes with everything.

Z(D4) contains R180 • R0 R90 R180 R270 H V D D'

Z(D4) = {R0, R180} • R0 R90 R180 R270 H V D D'

Z(G)≤G Proof: We will use the two step test. ex = xe for all x in G, so Z(G) is not empty. Choose any a and b in Z(G). Then for any x in G, we have (ab)x = a(xb) since b in Z(G) = (xa)b since a in Z(G) = x(ab) So Z(G) is closed.

Proof that Z(G) ≤ G (con't) To show Z(G) is closed under inverses, Choose any a in Z(G). For any x in G, ax = xa. Multiply on both sides by a-1: a-1 (ax)a-1 = a-1(xa)a-1 (a-1 a)(xa-1) = (a-1x)(aa-1) exa-1= a-1xe xa-1= a-1x By the two step test, Z(G) ≤ G

Centralizers Definition: Let a be any element of a group G. The centralizer of a in G, denoted C(a), is the set of elements that commutes with a. That is,

C(H) in D4 • R0 R90 R180 R270 H V D D' Start with row and column with H. See if Hx = xH

In D4, C(H) = {R0, R180, H, V} • R0 R90 R180 R270 H V D D'

Prove: C(a) ≤ G Proof: Let a be an element of a group G. We will use the one-step test to show that C(a) is a subgroup. ea = ae, so e belongs to C(a). Hence C(a) is nonempty.

Show xy-1 in C(a) Choose any x,y in C(a). Then (xy-1)-1a(xy-1) = (yx -1)a(xy-1) by S&S =yx-1(ax)y-1 = yx-1(xa)y-1 since x in C(a) =yay-1 since x-1x = e =(ay)y-1 since y is in C(a) = a since yy-1=e Multiply both sides on the left by (xy-1) to get: a(xy-1) = (xy-1)a Hence (xy-1) is in C(a) as required.

Subgroups of D4 {R0,R90,R180,R270,H,V,D,D'} {R0,R90,R180,R270} {R0,R180,D,D'} Trivial group Cyclic subgroups; In this case the Center is cyclic Centralizers: One of them is cyclic The whole group {R0, R180} {R0, D} {R0, D'} {R0, V} {R0, H} {R0}