In the book of Fundamental Concepts of Abstract Algebra by G. Ehrlich, at page 106, it is given that
Let $G$ be a group with centre $C_G$, and let $I_G$ be the group of all inner automorphisms of $G$. Then $I_g$ is isomorphic to $G/C_G$.
I have done its proof myself, and also the author provides another proof in the book. However, before reading the above theorem in the book, I had been trying to solve/understand the following question (asked in here) that I have come up with;
Let $G$ be a group, and $K \subseteq G$ be given. What are the necessary and sufficient condition for that there exists a normal subgroup $H$ of $G$ such that we can find a transversal $I$ of $G/H$ with with $I \subseteq K$.
, or alternatively a more weak result that (as a starting point),
Let $G$ be a group, and $K \subseteq G$ be given. Can we find an equivalence relation on $G$ such that there exists a transversal I of $G/K$ with $I \subseteq K$ ?
First of all, is there any intuitive way to understand what exactly does the theorem suggest ? For example, if we set $K = \{f_g : G\to G | f_g (x) = gxg^{-1}\}$, we have a set - even though we do not exactly know the elements of it - and we find that the distinct cosets of $C_G$ gives the distinct elements in $K$, which blew my mind at the fist reading. For example is there any "geometric" way of seeing this ?
Geometric way of seeing an Abstract thing is possibly impossible! But I have a short easy proof for your problem.
For a fixed $g\in G$, Define, $f:G\to I_G$ by $$f(g)=\phi_g, \forall g\in G$$ where $\phi_g:G\to G$ defined by $\phi_g(x)=gxg^{-1}$ for all $x\in G$.
Just check for $f$ to be a one-one homomorphism.
Next, find the kernel of $f$ i.e. $Ker(f)=\{g\in G:f(g)=I\}$, $I$ being the identity map. Precisely, $Ker(f)=Z(G)$, Center of $G$.
Hope this works.