Abstract Algebra – Why Isn’t This Free Product of Groups Abelian?

Question

I'm trying to prove that the free group $A=A_1*A_2$, where $A_1, A_2\neq 1$ is not abelian. Following the hints below:

Let $x,y\in A_1*A_2$, where $x\neq y$.

Suppose now $A_1=F(S)$ and $A_2=F(T)$, where $S=\{\alpha_1,\ldots,\alpha_n\}$ and $T=\{\beta_1,\ldots\beta_m\}$

Let $x,y\in A_1*A_2$, where $x\neq y$, then we have the words

$x=\alpha_1^{n_1}\ldots\alpha_k^{n_k}$ and $y=\beta_1^{m_1}\ldots\beta_l^{m_l}$

Thus using the definition of the operation of the free products, we have

$x\cdot y=\alpha_1^{n_1}\ldots\alpha_k^{n_k}\beta_1^{m_1}\ldots\beta_l^{m_l}$

$y\cdot x=\beta_1^{m_1}\ldots\beta_l^{m_l}\alpha_1^{n_1}\ldots\alpha_k^{n_k}$

Am I correct so far?

I can't continue from that point, since $k$ and $l$ can be different.

Thanks

Answer 1

I could find a clue for this question as follows which should be verified independently.

If $a\in A$ and $b\in B$ are nontrivial elements in $A*B$, then $aba^{-1}b^{-1}$ has infinite order and so the above group is an infinite centerless group$^1$.

$1$. An introduction to the Theory of Groups by J.J.Rotman.

If this clue is useful for paving the way of any answer, I will delete it. :)