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) • R0
R90
R180
R270 H V D D'

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

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

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

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

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

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. 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}