# Thread: Conjugacy in finite groups

1. ## Conjugacy in finite groups

Let $A, B$ be finite groups with $H \subset A \cap B$ be subgroup.
Suppose there exists a finite number of elements $h_{1}, h_{2}, \ldots, h_{r} \in H$ such that $a \sim_{A} h_{1} \sim_{B} h_{2} \sim_{A} \ldots \sim_{A(B)} h_{r}$ where $a \in A$.

I know that $a, h_{1}, h_{2}, \ldots, h_{r}$ all have same order.
But it doesn't mean $a, h_{r}$ in the same conjugacy class.

Is there anyway to obtain $a \sim_{A} h_{r}$ or $a, h_{r}$ in the same conjugacy class without adding abelian or self-conjugate?

2. ## Re: Conjugacy in finite groups

The only way I can understand your hypothesis is to assume both A and B are subgroups of a group G. In this case, it is obvious that a is conjugate to $h_r$ in G since conjugacy is an equivalence relation. Assuming $a\in A\cap B$, are you asking under what conditions is a conjugate to $h_r$ in A?

3. ## Re: Conjugacy in finite groups

Yes, $G=A^{*}_{H} B$. $G$ is the free product of groups $A$ and $B$ amalgamating subgroup $H$.
I have $A, B$ are finite.
I have results where ${x}^{A} \cap =\emptyset, {x}^{B} \cap =\emptyset$, where $||x||=0, ||y|| \leq 1$.

TO SHOW ${x}^{G} \cap =\emptyset$, I prove it by contradiction.

Suppose ${x}^{G} \cap \neq \emptyset$, so I have $x \sim_{G} y^{k}$, where $k \in \mathbb{Z}$.
By Lemma, there exists a finite number of elements $h_{i} \in H$ such that $y^{k} \sim_{A} h_{1} \sim_{B} h_{2} \sim_{A} \ldots \sim_{A(B)} h_{r}=x$.

To show it by contradiction, I need to have $y^{k} \sim_{A} x$ or $y^{k} \sim_{B} x$. I just can't see how to reach this step.