Proof of Aut(G): ϕ(Z(G))= Z(G)

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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.
 
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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)?
 
mykayla10 said:

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]