A problem on free products

Question

If $G=A * B$ is the free product of two groups $A$ and $B$ and $g \in G-A$, then prove that $gAg^{-1} \cap A=1$.

We know $A \cap B=1$, so if we write $g=a_1b_1a_2b_2 \ldots a_nb_n$, does not give me sufficient road to go? How should I approach it?

Answer 1

An idea: use induction on the length of a reduced normal word $\;g\in G\setminus A\;$ . For example, if $\;\ell(g)=1\;$, then $\;g=b\in B\;$ , so that for all $\;a\in A\;$:

$$gag^{-1}=bab^{-1}=a'\in A\implies ba=a'b $$

which is impossible, so that in this case we certainly have $\;bAb^{-1}\cap A=1\;$ .

If $\;\ell(g)=2\;$ you have two cases: $\;g=ab\;$ or $\;g=ba\;$, and as above etc.

For the general case, one further idea: if we assume truth for $\;\ell(g)<n\;$ and we now have $\;g\in G\setminus A\;$ with $\;\ell(g)=n\;$ , then the word $\;g\;$ is of the form $\;b_1a_2\ldots b_n\;$, or $\;b_1a_2\ldots a_n\;$ , or $\;a_1b_2\ldots a_n\;$ , or $\;a_1b_2\ldots b_n\;$ . Check each case and apply the inductive hypothesis.

For example, suppose $\;g=a_1b_2\ldots a_n\in G\setminus A\;$. Then, if for some $\;a\in A\;$ there exists $\;a'\in A\;$ s.t.

$$gag^{-1}=a'\implies ga=a'g\iff a_1b_2\ldots a_n a=a'a_1b_2\ldots a_n$$

Observe now that the length of both words in the right expression is $\;n+1\;$, so either they both are normal reduced and then $\;a'a_1=a_1\iff a'=1\;$ , which then gives

$$gag^{-1}=1\implies a=1$$

and we're done, or else $\;a_na=a_n\iff a=1\;$ and again we're done.

Thus, the words are unreduced and we can then reduce them. Try to take it from here.

Answer 2

Here is a sketch of proof based on algebraic topology:

Let $X_A$ and $X_B$ be two CW-complexes whose fundamental groups are $A$ and $B$ respectively. Choose two basepoints $x_A \in X_A$ and $x_B \in X_B$, and define $X$ as the complex obtained by attaching an edge between $X_A$ and $X_B$ at $x_A$ and $x_B$. In particular, $\pi_1(X) \simeq A \ast B$. The universal cover $\widetilde{X}$ can be described as a bipartite tree whose vertices are copies of $\widetilde{X_A}$ and $\widetilde{X_B}$. Let us consider the action of $A \ast B$ on $X$ by deck transformations. Notice that any element $h$ of $A \cap A^g$ stabilizes the copies $\widetilde{X_A}$ and $g \cdot \widetilde{X_A}$, with $g \cdot \widetilde{X_A} \neq \widetilde{X_A}$ whenever $g \notin A$. Therefore, $h$ stabilizes the geodesic between these two vertices of our tree; in particular, it fixes the edges. Because the action is free, we deduce that $h=1$.