Center of a free product

Let $A$ and $B$ be two groups and let $G=A*B$ be their free product. Suppose both $A$ and $B$ are nontrivial. I want now to prove that $G$ is infinite and that $Z(G)$ is trivial.

I think I can prove that $G$ is infinite using this argument: two reduced words of $A*B$ are equal if and only if they have the same length and the ith letter of the first word is the same as the ith letter of the second one. Hence one can find a reduced word of $G$ of length arbitrarily big (since we are sure it won't be equal to a shorter word). Do you think it works? Has anyone another suggestion?

On the other hand, I don't really have a clue for the second part except trying to analyze a long list of cases of reduced words and prove for each one that the only element with which they commute is the identity. Does anyone have a smarter idea?

Thanks in advance!

Well, you would have a hard time proving that a reduced word only commutes with the identity, because every word commutes with every power of itself, and of course, any element $a\in A$ will commute with all elements in $C_A(a)$, and any $b\in B$ with all elements in $C_B(b)$. So no element of $A*B$ has trivial centralizer! But you don't need a nontrivial element that doesn't commute with anything for the center to be trivial (good thing, since there is no such thing in any nontrivial group); you just need every nontrivial element to not commute with something.

You don't really need to consider a lot of cases: the only two things you have to worry a bit about are what to do if both of your groups have only two elements, and what the reduced word starts and ends with. For example, if $\mathbf{w}$ is a reduced word that starts and ends with an element of $A$, then just take $b\in B$, $b\neq e$ to get something it does not commute with, so $\mathbf{w}$ cannot be central. If it starts with an element of $A$ and ends with an element of $B$, then take a word $\mathbf{w}'$ that starts with an element of $A$ and ends with an element of $B$ (so that $\mathbf{w}\mathbf{w}'$ and $\mathbf{w}'\mathbf{w}$ are already reduced), and make $\mathbf{w}'$ either begin with something different from what $\mathbf{w}$ begins with, or end with something different from what $\mathbf{w}$ ends with; etc. You should have just 4 cases, and only two that can give you trouble if your groups are too small, in which case you can just stare at them directly.

Your argument for why $A*B$ is infinite is fine. You can even produce, explicitly, words of arbitrary length: just take $a\in A$, $b\in B$ with $a\neq e\neq b$, and consider $(ab)^n$, $n\in\mathbb{Z}$.

This is a slightly shorter version of Arturo Magidin's idea.

Theorem

If $A > 1$ and $B > 1$ are groups, then $Z(A * B) = 1$.

Proof

$A>1$ implies there exists $a \in A$ s.t. $a \neq e_A$.

$B>1$ implies there exists $b \in B$ s.t. $b \neq e_B$.

Let $w \in A*B$ be any nonempty reduced word. WLOG, $w = w_1\ldots w_k a_r$, where $a_r$ is a nonempty reduced word from $A$.

Then $bw_1\ldots w_k a_r \neq w_1\ldots w_k a_r b$ since the former, after reduction, must end in $a_r$; but the latter, after reduction, must not end in $a_r$! (Note that $a_r$ and $b$ cannot reduce with eachother because they come from separate groups)

Therefore $bw \neq wb$, and so $w \not\in Z(A*B)$.

Since $w$ was an arbitrary nonempty reduced word of $A*B$, it follows that only the empty word $e \in Z(A*B)$.

$Z(A*B) = 1$.