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

Then do the part involving h.

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

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