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Math Help - Conjugate subgroups

  1. #1
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    Conjugate subgroups

    Let G be a group and let g be some element of G. Given that \tau is an isomorphism is such that \tau(g) = g_1, show that S(g) and S(g_1) are conjugate subgroups of G. Here, S(x) is the subgroup of the set of automorphisms of G that fix the element x. In other words, S(x) = \{\sigma\in Aut G \bigr| \ \sigma(x) = x\}.

    I haven't gotten anywhere with this after about 5 hours now. I feel like the key to the problem will be pulling in \tau somehow- intuitively, g and g1 seem to me sort of like inverses of each other across the \tau isomorphism. Since \tau is a member of the group Aut(G), it has an inverse, say \tau^{-1} such that \tau^{-1}(g_1) = g and I feel like \tau somehow will be very involved in the conjugation of S(g). I don't know. I've been staring at a whiteboard basically writing the definition of conjugation over and over and over and over without getting anywhere. I already have proved that S(g)=\{\sigma\in Aut G \bigr| \ \sigma(g) = g\} is a subgroup of Aut(G). I'm stuck :/
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  2. #2
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    Re: Conjugate subgroups

    Quote Originally Posted by cubejunkies View Post
    Let G be a group and let g be some element of G. Given that \tau is an isomorphism is such that \tau(g) = g_1, show that S(g) and S(g_1) are conjugate subgroups of G. Here, S(x) is the subgroup of the set of automorphisms of G that fix the element x. In other words, S(x) = \{\sigma\in Aut G \bigr| \ \sigma(x) = x\}.

    I haven't gotten anywhere with this after about 5 hours now. I feel like the key to the problem will be pulling in \tau somehow- intuitively, g and g1 seem to me sort of like inverses of each other across the \tau isomorphism. Since \tau is a member of the group Aut(G), it has an inverse, say \tau^{-1} such that \tau^{-1}(g_1) = g and I feel like \tau somehow will be very involved in the conjugation of S(g). I don't know. I've been staring at a whiteboard basically writing the definition of conjugation over and over and over and over without getting anywhere. I already have proved that S(g)=\{\sigma\in Aut G \bigr| \ \sigma(g) = g\} is a subgroup of Aut(G). I'm stuck :/
    If you conjugate an element of S(g) by \tau^{-1}, you get an element of S(g_1). Similarly, conjugating an element of S(g_1) by \tau gives an element of S(g). Conjugation by a fixed element is a 1-1 mapping and the result follows.
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  3. #3
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    Re: Conjugate subgroups

    Hi,
    You really should be more careful in your posting of a question. It should read something like:

    Let $G$ be a group and $g\in G$. Define the stabilizer $S(g)$ of $g$ in $\text{Aut}(G)$ by $S(g)=\{\sigma\in\text{Aut}(G)\,:\,\sigma(g)=g\}$. For any $\tau\in\text{Aut}(G),\,\,S(\tau(g))=\tau S(g)\tau^{-1}$. That is, $S(g)$ and $S(\tau(g))$ are conjugate subgroups of $\text{Aut}(G)$.

    The proof is almost trivial: Let $\sigma\in S(\tau(g))$ Then $\tau(g)=\sigma(\tau(g))$ and so $g=(\tau^{-1}\sigma\tau)(g)$ or $\tau^{-1}\sigma\tau\in S(g)$. That is $\sigma=\tau(\tau^{-1}\sigma\tau)\tau^{-1}\in\tau S(g)\tau^{-1}$. Conversely, let $\sigma\in\tau S(g)\tau^{-1}$. Then $\tau^{-1}\sigma\tau\in S(g)$ or $\tau^{-1}\sigma\tau(g)=g$. Thus $\sigma(\tau(g))=\tau(g)$ and $\sigma \in S(\tau(g))$.

    The same proof shows that in any permutation group G acting on a set $\Omega$, the stabilizer subgroups of $i$ and $g(i)$ are conjugate for any $i\in\Omega,\,\,g\in G$.

    Personal lament. In the "old" days we wrote functions on the right and frequently as exponents. To me, the above proof is much easier with this notation. $i^{gh}=i^g$ if and only if $i^{ghg^{-1}}=i$
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