Proof of Aut(G): ϕ(Z(G))= Z(G)

[mykayla10]

Homework Statement

For every ϕ in Aut(G), ϕ(Z(G))= Z(G).

Homework Equations

Z(G):={g in G| gh=hg for all h in G}

The Attempt at a Solution

I haven't made too much progress on this one. I know that if I let g be an element of Z(G) that I need to prove that For every ϕ(g) is also and element of Z(G), which means I need to prove that for every h in G ϕ(g)h=hϕ(g). I just do not know where to go from there. I also do not even know where to begin in proving that Z(G) is an element of ϕ(Z(G)) so that I can completely prove the equality.

 

[Dick]

You know that phi is an automorphism. Which means that there is an element j of G such that phi(j)=h. Can you use that to prove phi(g)h=hphi(g)?

 

[Stephen Tashi]

Homework Statement

For every ϕ in Aut(G), ϕ(Z(G))= Z(G).

Homework Equations

Z(G):={g in G| gh=hg for all h in G}

The Attempt at a Solution

I haven't made too much progress on this one. I know that if I let g be an element of Z(G) that I need to prove that For every ϕ(g) is also and element of Z(G), which means I need to prove that for every h in G ϕ(g)h=hϕ(g). I just do not know where to go from there.

Do you mean: I need to prove that [tex] \phi(Z(G)) \subseteq Z(G) [/tex]?
That involves proving for each [tex] x \in G[/tex], [tex] x \in \phi(Z(G)) [/tex] implies [tex] x \in Z(G) [/tex].
Let [tex] x \in \phi(Z(G)) [/tex] Then there exists a [tex] g \in Z(G) [/tex] such that [tex] x = \phi(g) [/tex] because [tex] x [/tex] is in the image of [tex] Z(G) [/tex] under the mapping [tex] \phi [/tex]. (Now you have your [tex] \phi(g) [/tex] to work with.)

Then do the part involving [tex] h [/tex].

Since [tex] \phi [/tex] s an automorphism, there exists an element [tex] r [/tex] such that [tex] \phi^{-1}(h) = r [/tex]. Show [tex] gr = rg [/tex]. Then look at [tex] \phi(rg) = \phi(gr) [/tex]

I also do not even know where to begin in proving that Z(G) is an element of ϕ(Z(G)) so that I can completely prove the equality.

You mean "is a subset".

Let [tex] g \in Z(G)[/tex]. Look at [tex] x = \phi^{-1}(g) [/tex] Show [tex] x [/tex] commutes with all elements [tex] r \in G [/tex] That proves that [tex] g [/tex] is the image of an element in [tex] Z(G) [/tex] , so [tex] g \in \phi(Z(G)) [/tex]