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Centers and Centralizers

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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Centers and Centralizers

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  1. Centers and Centralizers

  2. 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,

  3. Z(D4)

  4. Z(D4) contains R0

  5. Z(D4) contains R180

  6. Z(D4) = {R0, R180}

  7. 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.

  8. 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

  9. 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,

  10. C(H) in D4

  11. In D4, C(H) = {R0, R180, H, V}

  12. 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.

  13. 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.

  14. {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}

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