Center is a normal subgroup of G 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.

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.

