Question

5. Prove that the center of a group is the intersection of all the centralizers. 6. Is the center of a group abelian? For any group element g is the centralizer of g abelian? 7. Prove that a group of even order must have an element x so that x= 2.

Answer 1

3 Let G be the group and Z(G) be the center of the group G, and ecg) be the be centralizer of an element of EG.. so let c= n clg) YEG Now, suppose REZCG) 5 a = ax e G + + ae G, 3 e ca) - xen cog) ge G7 - NEC so, Z(G) CC again let xEC v e GT , че е (8) a & g EG ng=ga >> MEZ(67) so, e CZ (G) so, Z(G)=c=0 clg) GEG

6 Yes, the center of a group is abelian. Z(G) = 2 REGi ng=ga Vg EGY not ay EZ(G), then ag=gn & EG and as y EZ(G) =) YEG so, ayyn & Hy €Z (G). So, Z(G) is abelian. No, centralizen of any group element g is not abelian. counter example: Let G=P4, then R180 € Da. here C (R18o) = Da which is not abelian

at sa at Giatal ,T-2 a EG a= a-1} then s and T are complement of each other. Tis nmempty as e-el, and T contains all clements of ander a Cother than identity). Naw, net nes, then n al - x1 + (x-1)-1 ales so, if nes, then no es so, the number of elements of s is even. Now, as all is even so, the number of elements of T is also even, and as eft, so there exists atleast one corodd number of element(s) of order 2. so, a group of even onder must have an element x, so that 1x1=2 .