(Sorry for the long read, you may skip to the very end, to see the question there.)
I am reading an article and trying to understand the proof, but cannot work out a crucial detail. Here's the statement:
Theorem. Let $\Gamma$ be a subgroup of finite index in Mod(S); and let $\Gamma'$ be a torsionless group containing $\Gamma$ as a subgroup of finite index. Then $\Gamma'$ is naturally contained in Mod(S).
Here $S$ is a compact orientable surface, possibly with boundary, of genus at least 2, which is not a closed surface of genus 2, and Mod(S) is the extended mapping class group. The proof is very short and goes as follows:
We may assume that $\Gamma$ is normal in $\Gamma'$ and centerless, replacing, if necessary, $\Gamma$ by a smaller subgroup. Then the action of $\Gamma'$ on $\Gamma$ by conjugation induces a map $\Gamma'\to Aut(\Gamma)$. The facts that $\Gamma'$ is torsionless, $\Gamma$ is of finite index in $\Gamma'$, and $\Gamma$ is centerless imply that this map is injective. On the other hand, it follows from Theorem 2 that $Aut(\Gamma)$ is naturally contained in Mod(S).
I totally understand why we may assume that $\Gamma$ is normal in $\Gamma'$. Indeed, the intersection of all conjugates of $\Gamma$ in Mod(S) is also of finite index and is normal (the so-called core of $\Gamma$). However, I do not understand why we may assume that $\Gamma$ has trivial center. If we assume this, we can prove injectivity as follows. Take an element $g$ from the kernel of the map $\Gamma'\to Aut(\Gamma)$. Then $g$ centralizes every element of $\Gamma$, and some power $g^N$ of $g$ lies in $\Gamma$, since $\Gamma$ has finite index in $\Gamma'$. This means that $g^N$ lies in the center of $\Gamma$ which we assumed to be trivial. Hence $g^N=1$ and since $\Gamma'$ is torsionless, $g=1$.
The only problem I have is to see why we can assume that $\Gamma$ is centerless. Any hints?
UPDATE: Obviously, if a surface has more than one boundary component then it has a central direct factor isomorphic to $\mathbb Z$, and the statement is false. One explanation to that obvious discrepancy is that it may be that the author considers mapping classes up to isotopies which are allowed to move the boundary (it is not stated explicitly in the article), which is essentially equivalent to working with surfaces with punctures, not with boundary. With such interpretation the question becomes:
Suppose $S$ is an orientable surface, possibly with punctures, of genus at least 2, which is not the closed surface of genus 2. It is known that Mod(S) is centerless (Farb-Margalit, section 3.5). How do we prove that for any finite index subgroup $\Gamma$ in Mod(S) there is a finite index subgroup $\Gamma_1$ of $\Gamma$ which is centerless and normal?