$G$ is a group and $Z(G)$ its center. $f\colon G\to G$ is an automorphism of $G$. Show that if $x$ is in $Z(G)$, then $f(x)$ is also in $Z(G)$. $G$ is a group and $Z(G)$ its center. $f\colon G\to  G$ is an automorphism of $G$. Show that if $x$ is in $Z(G)$, then $f(x)$ is also in $Z(G)$.
So $x$ commutes with every element in $G$, and since $f$ is an automorphism of $G$, then $f(x)=x$ is in $Z(G)$. Can I prove like this? 
 A: No, for in general $f(x)\ne x$. To show that $f(x)\in Z(G)$ you need to show $f(x)y=yf(x)$ for all $y\in G$. Use the fact that wou can find $z$ with $y=f(z)$ to proceed.
A: It need not be the case that $f(x) = x$.  However, if $x$ is in the centre of the group, then $xy = yx$, for all $y$.  Now just apply $f$ to both sides of this equation, and use the fact that $f$ is a homomorphism, and onto.
A: The Exercise
Brah,  $Z(G)=\{x\in G: gx=xg,$ $ \forall g\in G\}$. Take an element $x\in Z(G)$ and ask the question "$f(x)f(g)=f(g)f(x)$ for all $g\in G$ ", if the answer is yes $f(x)\in Z(G)$(since $f$ is surjective ie for every $y\in G$ there exists a $y'\in G$ such that $f(y')=y$).

Automorphisms
Also an automorphism is just an isomorphism from the group to itself. Here is an example: Take an element $a\in G$ and define $f_a:G\to G$ where $f_a(x)=axa^{-1}$. it follows that:

*

*$f_a(xy)=axya^{-1}=axa^{-1}aya^{-1}=f_a(x)f_a(y)$

*$\forall x,y\in G$ it is clear that $axa^{-1}=aya^{-1}\implies x=y$ so the $\ker f=\{e\}$ which implies that $f$ is injective.

*$\forall x\in G $ it follows that $ax^{-1}a^{-1}=y^{-1}$ for some $y\in G$ so it is clear that $x^{-1}=a^{-1}y^{-1}a\implies x=aya^{-1}$, so $f$ is surjective.

In this example you can see that automorpisms don't have to be the identity map $f(x)=x$.
Also in this example you can see that:
If $x\in Z(G)$ then $f_a(x)f_a(g)=f_a(xg)=f_a(gx)=f_a(g)f_a(x)$ for all $g\in G$ so $f_a(x)\in Z(G)$ which makes since because $axa^{-1}=aa^{-1}x=x$ since $x\in Z(G)$
