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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(D 4 ). Z(D 4 ) contains R 0. Z(D 4 ) contains R 180. Z(D 4 ). = {R 0 , R 180 }. Z(G)≤G.
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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, R180}
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,
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.
{R0,R90,R180,R270} {R0,R180,H,V} {R0,R180,D,D'} {R0, H} {R0, V} {R0, R180} {R0, D} {R0, D'} Subgroups of D4 {R0,R90,R180,R270,H,V,D,D'} {R0}