TITLE: Center is a normal subgroup of G
QUESTION [5 upvotes]: This is a problem from Herstein's Topics in Algebra. 
I have already shown the above result using the definition of normal subgroup. But now I want to prove it by constructing a homomorphism such that kernel is center of the group G. 
How can I construct such homomorphism?
I was thinking of going like this. Given a $g$ in $G$ construct $E_g(x)=g x g^{-1}$
So given each element we have a transformation. Set of transformations like this form a group with inverse given by $E_{g^{-1}}$.
Kernel consist of all those elements for which $E_g$ is identity. In other words, $E_g(x)=x$ or $g x g^{-1}=x$ or $gx=xg $ for all $x$. That is $g$ commutes with everything in $G$. 
Am I going into the right direction.
Edit: I became interested in proving the result through homomorphism approach as problem is in section 2.7 which is titled homomorphisms. Herstein must be expecting us to take this route.

REPLY [2 votes]: You are very close. You already have a way to transform $g$ into an element of... something ... where $g$ transforms into the identity function if and only if $g$ communtes with everything in $G$. You also know that the something contains some sort of mappings.
Now you need to write that down with correct terms. That is, instead of "transforming $g$ into an element of some set that contains mappings, you need to 

Write down exactly what set $g$ is mapping into, what the operation on that set is, and make sure that set is a group.
Write down the mapping as a mapping between two groups
Prove that the mapping is a homomorphism.