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,

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Z(G) = {g∈ G | gx = xg for all x ∈ G}

Z(D4)

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

Z(D4) contains R0

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

Z(D4) contains R180

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

Z(D4)

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

= {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,

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C(a) = {g∈ G | ga = ag}

C(H) in D4

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

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

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

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, R180} {R0, D} {R0, D'}{R0, V}{R0, H}

{R0,R90,R180,R270}{R0,R180,H,V} {R0,R180,D,D'}

{R0,R90,R180,R270,H,V,D,D'}

{R0}