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Thread: Conjugacy in finite groups

  1. #1
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    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?
    Last edited by deniselim17; Sep 26th 2017 at 05:42 AM.
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  2. #2
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    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?
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  3. #3
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    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 <y> =\emptyset, {x}^{B} \cap <y> =\emptyset, where ||x||=0, ||y|| \leq 1.

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

    Suppose {x}^{G} \cap <y> \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.
    Last edited by deniselim17; Sep 26th 2017 at 06:15 PM.
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