Center is a normal subgroup of G

Question

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.

Answer 1

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

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