I am currently reading through a solution of the following problem and I am having a bit of trouble. The problem is as follows:
Let $S$ be a set and $F$ be the free group on $S$. Show that if $|S| > 1$, then $Z(F)$ is trivial.
In order to deduce a contradiction, suppose that $w \in Z(F) \setminus \{e\}$ (where $e$ is the identity element). Let $w = a_{1}^{\pm1} \cdots a_{n}^{\pm 1}$ be the reduced word expression for $w$. Since $w \in Z(F)$, we have $$ \left(a_{n-1}^{\mp 1} \cdots a_{1}^{\mp 1} \right)w = w\left(a_{n-1}^{\mp 1} \cdots a_{1}^{\mp 1} \right). $$ Now the left-hand side reduces to $$ \left(a_{n-1}^{\mp 1} \cdots a_{1}^{\mp 1} \right)w = \left(a_{n-1}^{\mp 1} \cdots a_{1}^{\mp 1} \right)\left(a_{1}^{\pm1} \cdots a_{n}^{\pm 1} \right) = a_{n}^{ \pm 1}. $$ Comparing this to the right-hand side, we deduce that it must be the case that $a_{n}^{\pm 1}=a_{n-1}^{\pm 1} = \cdots = a_{1}^{\pm 1}$; so write $w = \left(a_{1}^{\pm 1}\right)^n$.
[Now every part of the proof makes sense to me thus far, but I do not see the rationale for the coming part.]
Since $|S| > 1$, let $b \in S$ be such that $a_{1}^{\pm 1} \neq b$. Then $wb = bw$ because $w \in Z(F)$, and so $\left( a_{1}^{\pm1} \right)^n b = b\left(a_{1}^{\pm 1} \right)^n$. However this implies that $a_{1}^{\pm 1} = b$, which is a contradiction.
I cannot see how to deduce that $a_{1}^{\pm 1} = b$ from $\left( a_{1}^{\pm1} \right)^n b = b\left(a_{1}^{\pm 1} \right)^n$. The solution says to consider the reduced words, but this does not seem to be very helpful considering $\left( a_{1}^{\pm1} \right)^n b$ and $b\left( a_{1}^{\pm1} \right)^n$ are already reduced.
