Centralizer of factor group I'm dealing with a finite $p$-group $G$ whose derived subgroup $G'$ is cyclic. The author defines $C^*\leq G$ as
\begin{equation*}
 C^*=C_G\big(G'/(G')^{p^2}\big).
\end{equation*}
By definition, the centralizer $C_G(S)$ of a subset $S$ of $G$ is the subgroup given by all elements $g$ of $G$ such that $[g,s]=1\, \forall s\in S$. I'm wondering how I can define a subgroup like $C^*$ as I'm being asked to compute the commutator between an element of $G$ and a coset in $G'/(G')^{p^2}$. 
The full article can be read here, this centralizer can be found at page 113.
 A: There are two ways to interpret this, both leading to the same set of elements.
One is, as the_fox mentions, to understand this as the set of elements in $G$ that map, under the canonical map $G\to G/(G')^{p^2}$, to the centralizer of $G'/(G')^{p^2}$. This is seen in the definition of the second center, which is sometimes said to be the centralizer in $G$ of $G/Z(G)$ (well, not literally that, as you note).
The other way to interpret this is that $G$ has a natural action on $G'/(G')^{p^2}$ by conjugation, and you are asking for the centralizer in that action, that is, the elements that fix everything: if $G$ is a group acting on the left on $H$, then for $A\leq H$ we let $C_G(A)=\{g\in G\mid {}^ga = a\text{ for all }a\in A\}$
In fact, the two notions coincide. For simplicity, I will let $(G')^{p^2}=N$.
Suppose $g\in G$ is such that $gN\in C_{G/N}(G'/N)$. That means that for every $x\in G'$, $(Nx)(Ng) = (Ng)(Nx)$, or equivalently, that $Ngxg^{-1} = Nx$. This means that ${}^g(Nc) = Nx$ for all $x\in G'$, so $g$ centralizes $G'/N$ under its standard action.
Conversely, if ${}^g(Nx)=Nx$ for all $x\in G'$, then $Ngxg^{-1}N=Nx$, so $(Ng)(Nx)=(Nx)(Ng)$ in $G/N$; that is, $gN\in C_{G/N}(G'/N)$. 
So the two interpretations result in the exact same subgroup of $G$. Of course you can replace $G'$ with some other subgroup containing $N$.  
