Let $G$ be a group with trivial centre. For $H \le G$, let $C_{G'}(G)=\{g' \in G': gg'=g'g, \forall g \in G\}$. This is my question: Given a group $G$, does there exist a non-trivial group $G'$ such that $C_{G'}(G)=\{e\}$ and $G \triangleleft G'$?
I can always find a group $G'$ such that $G \triangleleft G'$. Let $H$ be any nontrivial group in the world with identity $e$. Consider the Cartesian product $G \times H$. The product is a group with respect to the corresponding operations on the corresponding parts. It is easy to check that $G \cong G \times \{e\} $. I claim that $G$ is normal in $G'=G \times H$. To see this observe that for any $(g,h) \in G'$, we have $(g,h)(g_1,e)(g^{-1},h^{-1})=(gg_1g^{-1},e) $.
This method won't work to prove the second condition. Because $(e_G,H) \subset C_{G'}(G)$ in my construction. How do I construct the bigger group, if at all it is possible?
Just a hint would suffice. Thanks for the help!!
Since the center of $G$ is trivial, $G\cong \mathrm{Inn}(G)\lhd \mathrm{Aut}(G)$, so $G\rtimes G$ will work fine.
Note that the construction @Berci suggests only gives the desired property if the homomorphism $H\to \mathrm{Aut}(G)$ is injective.