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Thread: Conjugates in a finite group

  1. #1
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    Conjugates in a finite group

    Show that in a finite group of order n, an element of order k has at most n/k conjugates.

    Proof so far.

    Let G be a group with order n, and suppose that g is an element with the order of k. Then I have k|n by the property of order. But I'm not too familiar in dealing with conjugates, how should I get started here? Thanks.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Show that in a finite group of order n, an element of order k has at most n/k conjugates.
    Let $\displaystyle G$ act on itself by conjugative. Then the conjugates to $\displaystyle a\in G$ is the orbit $\displaystyle Ga$. But $\displaystyle |Ga| = |G:G_a|$. And $\displaystyle |G_a| = |C(a)|$ - the centralizer. But $\displaystyle \{ a,a^2,...,a^k\} \subseteq C(a)$. Therefore, $\displaystyle |C(a)| \geq k$. This means $\displaystyle [G:G_a] \leq n/k$. Therefore, $\displaystyle |Ga| \leq n/k$.
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