# Answer to Question #6226 in Abstract Algebra for Lakeisha Lewis

Question #6226

Z(G)={Z in G/ zx=xz for all x in G} write a proof to show z(G) is a subgroup of G

Expert's answer

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

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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