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Math Help - Conjugacy class

  1. #1
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    Conjugacy class

    I'm stuck on this problem. Hope someone can give me help.
    Let H be a normal subgroup of G, show that if x \in H then ord(x^H)=ord(x^G) or ord(x^H)=\frac{1}{2}ord(x^G)
    I think I need to construct an isomorphism, but I have no clue how to do it here.
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  2. #2
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    Quote Originally Posted by jackie View Post
    I'm stuck on this problem. Hope someone can give me help.
    Let H be a normal subgroup of G, show that if x \in H then ord(x^H)=ord(x^G) or ord(x^H)=\frac{1}{2}ord(x^G)
    I think I need to construct an isomorphism, but I have no clue how to do it here.

    What is x^H? Is this the normal closure of x (of <x>) in H? Wouldn't ord(x^H)=\frac{1}{2}ord(x^H) mean that |H| has to be even?
    Where did you take this problem from?

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    What is x^H? Is this the normal closure of x (of <x>) in H? Wouldn't ord(x^H)=\frac{1}{2}ord(x^H) mean that |H| has to be even?
    Where did you take this problem from?

    Tonio
    G and H act on itself by conjugation
    I think x^H=\{gxg^{-1}:g \in H\}={conjugates of x in H}
    and similarly for x^G. Also, G is finite
    This problem is from our homework handout
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  4. #4
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    Quote Originally Posted by jackie View Post
    G and H act on itself by conjugation
    I think x^H=\{gxg^{-1}:g \in H\}={conjugates of x in H}
    and similarly for x^G. Also, G is finite
    This problem is from our homework handout

    I think it must be x^H=<gxg^{-1}; g \in H> meaning: the subgroup generated by the conjugates of x by elements of H = the normal closure of <x> in H.

    As H \vartriangleleft G, both sbgps. x^H \mbox { and } x^G are contained in H

    Tonio
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    Quote Originally Posted by jackie View Post
    I'm stuck on this problem. Hope someone can give me help.
    Let H be a normal subgroup of G, show that if x \in H then ord(x^H)=ord(x^G) or ord(x^H)=\frac{1}{2}ord(x^G)
    I think I need to construct an isomorphism, but I have no clue how to do it here.
    this claim is false! (unless by x^G and x^H you mean something else!) here's a counter-example:

    let p \geq 5 be a prime number and define G=\left \{\begin{pmatrix}a & b \\ 0 & c \end{pmatrix}: \ \ a,b,c \in \mathbb{Z}/p\mathbb{Z}, \ ac \neq 0 \right \} and H=\left \{\begin{pmatrix}1 & a \\ 0 & 1 \end{pmatrix}: \ \ a \in \mathbb{Z}/p\mathbb{Z} \right \}. also choose x=\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}.

    see that H is a normal subgroup of G. a little work shows that |x^H|=|\{hxh^{-1}: \ h \in H \}|=1 and |x^G|=|\{gxg^{-1}: \ g \in G \}|=p-1.

    however, it can be shown that for any finite group G, any normal subgroup H of G, and any x \in H we have either |x^G|=|x^H| or |x^G| \geq 2|x^H|.
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  6. #6
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    Thank you very much for the help, tonio and NonCommAlg.
    You are right NonCommAlg. The question was supposed to include that H is a normal subgroup of index 2 in finite G.
    I used the result [G:C_G(x)]=ord(x^G), similarly [H:C_H(x)]=ord(x^H). I also used the second isomorphism in my proof.
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