Z(G)={Z in G/ zx=xz for all x in G} write a proof to show z(G) is a subgroup of G
if a is in Z(G) then it inverse a^-1 also in Z(G)
ax=xa for every x. we can
multiply on the both side on a^-1 then we have
a^(-1)*a*x*a^(-1)=a^(-1)*x*a*a^(-1) so x*a^(-1)=a^(-1)*x
1 is from
Z(G)
associative rule is true because it is true in G
it is enough that
this subset is closed by multiplication
if a,b from Z(G) then ab is from
Z(G)
ax=xa for every x
by=yb for every y
so if y=ax=xa we have
b*y=b*a*x=y*b=x*a*b
so bax=xab but ba=ab from definition of Z(G)
then
ab*x=x*ab for every x
the close for operation subset of group is subgroup
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