Real Conjugacy Class, group theory












1












$begingroup$


Let G be a finite group of odd order n. Suppose that g is in a real conjugacy class C (a real conjugacy class C is a conjugacy class so that $C$ = $C^{-1}$ ). So $h^{−1}gh = g^{−1}$ for some h $in G$.



Why can we follow from that: $h^{−2}gh^2 = g$ ?



This means $h^2 in C_G(g)$. Since n is odd, the order of h is odd, say
2k + 1. It follows that h = ($h^2)^{k+1}$



Why can we follow from that: h $in C_G(g)$ ?



Thankfull for any help.






abstract-algebra group-theory finite-groups







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let G be a finite group of odd order n. Suppose that g is in a real conjugacy class C (a real conjugacy class C is a conjugacy class so that $C$ = $C^{-1}$ ). So $h^{−1}gh = g^{−1}$ for some h $in G$.



    Why can we follow from that: $h^{−2}gh^2 = g$ ?



    This means $h^2 in C_G(g)$. Since n is odd, the order of h is odd, say
    2k + 1. It follows that h = ($h^2)^{k+1}$



    Why can we follow from that: h $in C_G(g)$ ?



    Thankfull for any help.






    abstract-algebra group-theory finite-groups







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let G be a finite group of odd order n. Suppose that g is in a real conjugacy class C (a real conjugacy class C is a conjugacy class so that $C$ = $C^{-1}$ ). So $h^{−1}gh = g^{−1}$ for some h $in G$.



      Why can we follow from that: $h^{−2}gh^2 = g$ ?



      This means $h^2 in C_G(g)$. Since n is odd, the order of h is odd, say
      2k + 1. It follows that h = ($h^2)^{k+1}$



      Why can we follow from that: h $in C_G(g)$ ?



      Thankfull for any help.






      abstract-algebra group-theory finite-groups







      share|cite|improve this question











      $endgroup$




      Let G be a finite group of odd order n. Suppose that g is in a real conjugacy class C (a real conjugacy class C is a conjugacy class so that $C$ = $C^{-1}$ ). So $h^{−1}gh = g^{−1}$ for some h $in G$.



      Why can we follow from that: $h^{−2}gh^2 = g$ ?



      This means $h^2 in C_G(g)$. Since n is odd, the order of h is odd, say
      2k + 1. It follows that h = ($h^2)^{k+1}$



      Why can we follow from that: h $in C_G(g)$ ?



      Thankfull for any help.






      abstract-algebra group-theory finite-groups




      abstract-algebra group-theory finite-groups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 16:23







      FabianSchneider

















      asked Dec 12 '18 at 20:57









      FabianSchneiderFabianSchneider

      706




      706






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          The order of $|G|$ is odd, say $G=2m-1$, for some positive integer $m$. Hence by Lagrange $h^{|G|}=1=(h^2)^m cdot h^{-1}$. It follows that $h=(h^2)^m$, so if $h^2 in C_G(g)$, then $h in C_G(g)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you please be more precise at the end where you follow $h in C_G(g)$. That´s the part where I´m struggeling.
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:45












          • $begingroup$
            If $h^2$ is in the subgroup then any power of it is too.
            $endgroup$
            – Nicky Hekster
            Dec 12 '18 at 21:49










          • $begingroup$
            Makes sense. Thank you so much!
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:52



















          1












          $begingroup$

          Rewrite the equality as $h^{-1}g^{-1}h=g$ and now(here’s the trick!) replace the g in $h^{-1}g^{-1}h$ with $h^{-1}g^{-1}h$ (because g and $h^{-1}g^{-1}h$ are equal). So we get that $g=h^{-1}g^{-1}h=h^{-1}(h^{-1}g^{-1}h)^{-1}h=h^{-2}gh^2$ keep replacing g inside $h^{-2}gh^2$ with $h^{-1}g^{-1}h$ to eventually get that $h^{-2k}gh^{2k}=g$, but $h^{2k}=h^{-1}$ so by the last equality $h$ and $g$ commute






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we, after k-1 substitutions, have $ g = h^{-2k}gh^{2k}$ and not $ g = h^{-2k}g^{-1}h^{2k}$ ?
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:43








          • 1




            $begingroup$
            You get $g = h^{-2k}gh^{2k}$ after $2k $ substitutions, not after $ k$, and the reason you get $ g $ and not $ g^{-1}$ is because $ 2k $ is even and after an even number of substitutions you get g while after an odd number of substitutions you get $g^{-1}$
            $endgroup$
            – Sorin Tirc
            Dec 12 '18 at 23:19












          • $begingroup$
            I meant 2k-1 (not k-1) but i was still false....thank you for your help! I got it know.
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 23:24










          • $begingroup$
            We can´t follow $h^{-1}g^{-1}h=g$ from $h^{-1}gh = g^{-1}$, because G is not neccesarry commutative ?!
            $endgroup$
            – FabianSchneider
            Dec 13 '18 at 16:35








          • 1




            $begingroup$
            you are partly right because it should be $hg^{-1}h^{-1}=g$ but the argument still goes through ;)
            $endgroup$
            – Sorin Tirc
            Dec 13 '18 at 20:39













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          2 Answers
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          2 Answers
          2






          active

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          active

          oldest

          votes






          active

          oldest

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          1












          $begingroup$

          The order of $|G|$ is odd, say $G=2m-1$, for some positive integer $m$. Hence by Lagrange $h^{|G|}=1=(h^2)^m cdot h^{-1}$. It follows that $h=(h^2)^m$, so if $h^2 in C_G(g)$, then $h in C_G(g)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you please be more precise at the end where you follow $h in C_G(g)$. That´s the part where I´m struggeling.
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:45












          • $begingroup$
            If $h^2$ is in the subgroup then any power of it is too.
            $endgroup$
            – Nicky Hekster
            Dec 12 '18 at 21:49










          • $begingroup$
            Makes sense. Thank you so much!
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:52
















          1












          $begingroup$

          The order of $|G|$ is odd, say $G=2m-1$, for some positive integer $m$. Hence by Lagrange $h^{|G|}=1=(h^2)^m cdot h^{-1}$. It follows that $h=(h^2)^m$, so if $h^2 in C_G(g)$, then $h in C_G(g)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you please be more precise at the end where you follow $h in C_G(g)$. That´s the part where I´m struggeling.
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:45












          • $begingroup$
            If $h^2$ is in the subgroup then any power of it is too.
            $endgroup$
            – Nicky Hekster
            Dec 12 '18 at 21:49










          • $begingroup$
            Makes sense. Thank you so much!
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:52














          1












          1








          1





          $begingroup$

          The order of $|G|$ is odd, say $G=2m-1$, for some positive integer $m$. Hence by Lagrange $h^{|G|}=1=(h^2)^m cdot h^{-1}$. It follows that $h=(h^2)^m$, so if $h^2 in C_G(g)$, then $h in C_G(g)$.






          share|cite|improve this answer









          $endgroup$



          The order of $|G|$ is odd, say $G=2m-1$, for some positive integer $m$. Hence by Lagrange $h^{|G|}=1=(h^2)^m cdot h^{-1}$. It follows that $h=(h^2)^m$, so if $h^2 in C_G(g)$, then $h in C_G(g)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 21:17









          Nicky HeksterNicky Hekster

          28.4k53456




          28.4k53456












          • $begingroup$
            Can you please be more precise at the end where you follow $h in C_G(g)$. That´s the part where I´m struggeling.
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:45












          • $begingroup$
            If $h^2$ is in the subgroup then any power of it is too.
            $endgroup$
            – Nicky Hekster
            Dec 12 '18 at 21:49










          • $begingroup$
            Makes sense. Thank you so much!
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:52


















          • $begingroup$
            Can you please be more precise at the end where you follow $h in C_G(g)$. That´s the part where I´m struggeling.
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:45












          • $begingroup$
            If $h^2$ is in the subgroup then any power of it is too.
            $endgroup$
            – Nicky Hekster
            Dec 12 '18 at 21:49










          • $begingroup$
            Makes sense. Thank you so much!
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:52
















          $begingroup$
          Can you please be more precise at the end where you follow $h in C_G(g)$. That´s the part where I´m struggeling.
          $endgroup$
          – FabianSchneider
          Dec 12 '18 at 21:45






          $begingroup$
          Can you please be more precise at the end where you follow $h in C_G(g)$. That´s the part where I´m struggeling.
          $endgroup$
          – FabianSchneider
          Dec 12 '18 at 21:45














          $begingroup$
          If $h^2$ is in the subgroup then any power of it is too.
          $endgroup$
          – Nicky Hekster
          Dec 12 '18 at 21:49




          $begingroup$
          If $h^2$ is in the subgroup then any power of it is too.
          $endgroup$
          – Nicky Hekster
          Dec 12 '18 at 21:49












          $begingroup$
          Makes sense. Thank you so much!
          $endgroup$
          – FabianSchneider
          Dec 12 '18 at 21:52




          $begingroup$
          Makes sense. Thank you so much!
          $endgroup$
          – FabianSchneider
          Dec 12 '18 at 21:52











          1












          $begingroup$

          Rewrite the equality as $h^{-1}g^{-1}h=g$ and now(here’s the trick!) replace the g in $h^{-1}g^{-1}h$ with $h^{-1}g^{-1}h$ (because g and $h^{-1}g^{-1}h$ are equal). So we get that $g=h^{-1}g^{-1}h=h^{-1}(h^{-1}g^{-1}h)^{-1}h=h^{-2}gh^2$ keep replacing g inside $h^{-2}gh^2$ with $h^{-1}g^{-1}h$ to eventually get that $h^{-2k}gh^{2k}=g$, but $h^{2k}=h^{-1}$ so by the last equality $h$ and $g$ commute






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we, after k-1 substitutions, have $ g = h^{-2k}gh^{2k}$ and not $ g = h^{-2k}g^{-1}h^{2k}$ ?
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:43








          • 1




            $begingroup$
            You get $g = h^{-2k}gh^{2k}$ after $2k $ substitutions, not after $ k$, and the reason you get $ g $ and not $ g^{-1}$ is because $ 2k $ is even and after an even number of substitutions you get g while after an odd number of substitutions you get $g^{-1}$
            $endgroup$
            – Sorin Tirc
            Dec 12 '18 at 23:19












          • $begingroup$
            I meant 2k-1 (not k-1) but i was still false....thank you for your help! I got it know.
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 23:24










          • $begingroup$
            We can´t follow $h^{-1}g^{-1}h=g$ from $h^{-1}gh = g^{-1}$, because G is not neccesarry commutative ?!
            $endgroup$
            – FabianSchneider
            Dec 13 '18 at 16:35








          • 1




            $begingroup$
            you are partly right because it should be $hg^{-1}h^{-1}=g$ but the argument still goes through ;)
            $endgroup$
            – Sorin Tirc
            Dec 13 '18 at 20:39


















          1












          $begingroup$

          Rewrite the equality as $h^{-1}g^{-1}h=g$ and now(here’s the trick!) replace the g in $h^{-1}g^{-1}h$ with $h^{-1}g^{-1}h$ (because g and $h^{-1}g^{-1}h$ are equal). So we get that $g=h^{-1}g^{-1}h=h^{-1}(h^{-1}g^{-1}h)^{-1}h=h^{-2}gh^2$ keep replacing g inside $h^{-2}gh^2$ with $h^{-1}g^{-1}h$ to eventually get that $h^{-2k}gh^{2k}=g$, but $h^{2k}=h^{-1}$ so by the last equality $h$ and $g$ commute






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why do we, after k-1 substitutions, have $ g = h^{-2k}gh^{2k}$ and not $ g = h^{-2k}g^{-1}h^{2k}$ ?
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:43








          • 1




            $begingroup$
            You get $g = h^{-2k}gh^{2k}$ after $2k $ substitutions, not after $ k$, and the reason you get $ g $ and not $ g^{-1}$ is because $ 2k $ is even and after an even number of substitutions you get g while after an odd number of substitutions you get $g^{-1}$
            $endgroup$
            – Sorin Tirc
            Dec 12 '18 at 23:19












          • $begingroup$
            I meant 2k-1 (not k-1) but i was still false....thank you for your help! I got it know.
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 23:24










          • $begingroup$
            We can´t follow $h^{-1}g^{-1}h=g$ from $h^{-1}gh = g^{-1}$, because G is not neccesarry commutative ?!
            $endgroup$
            – FabianSchneider
            Dec 13 '18 at 16:35








          • 1




            $begingroup$
            you are partly right because it should be $hg^{-1}h^{-1}=g$ but the argument still goes through ;)
            $endgroup$
            – Sorin Tirc
            Dec 13 '18 at 20:39
















          1












          1








          1





          $begingroup$

          Rewrite the equality as $h^{-1}g^{-1}h=g$ and now(here’s the trick!) replace the g in $h^{-1}g^{-1}h$ with $h^{-1}g^{-1}h$ (because g and $h^{-1}g^{-1}h$ are equal). So we get that $g=h^{-1}g^{-1}h=h^{-1}(h^{-1}g^{-1}h)^{-1}h=h^{-2}gh^2$ keep replacing g inside $h^{-2}gh^2$ with $h^{-1}g^{-1}h$ to eventually get that $h^{-2k}gh^{2k}=g$, but $h^{2k}=h^{-1}$ so by the last equality $h$ and $g$ commute






          share|cite|improve this answer









          $endgroup$



          Rewrite the equality as $h^{-1}g^{-1}h=g$ and now(here’s the trick!) replace the g in $h^{-1}g^{-1}h$ with $h^{-1}g^{-1}h$ (because g and $h^{-1}g^{-1}h$ are equal). So we get that $g=h^{-1}g^{-1}h=h^{-1}(h^{-1}g^{-1}h)^{-1}h=h^{-2}gh^2$ keep replacing g inside $h^{-2}gh^2$ with $h^{-1}g^{-1}h$ to eventually get that $h^{-2k}gh^{2k}=g$, but $h^{2k}=h^{-1}$ so by the last equality $h$ and $g$ commute







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 21:17









          Sorin TircSorin Tirc

          1,755213




          1,755213












          • $begingroup$
            Why do we, after k-1 substitutions, have $ g = h^{-2k}gh^{2k}$ and not $ g = h^{-2k}g^{-1}h^{2k}$ ?
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:43








          • 1




            $begingroup$
            You get $g = h^{-2k}gh^{2k}$ after $2k $ substitutions, not after $ k$, and the reason you get $ g $ and not $ g^{-1}$ is because $ 2k $ is even and after an even number of substitutions you get g while after an odd number of substitutions you get $g^{-1}$
            $endgroup$
            – Sorin Tirc
            Dec 12 '18 at 23:19












          • $begingroup$
            I meant 2k-1 (not k-1) but i was still false....thank you for your help! I got it know.
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 23:24










          • $begingroup$
            We can´t follow $h^{-1}g^{-1}h=g$ from $h^{-1}gh = g^{-1}$, because G is not neccesarry commutative ?!
            $endgroup$
            – FabianSchneider
            Dec 13 '18 at 16:35








          • 1




            $begingroup$
            you are partly right because it should be $hg^{-1}h^{-1}=g$ but the argument still goes through ;)
            $endgroup$
            – Sorin Tirc
            Dec 13 '18 at 20:39




















          • $begingroup$
            Why do we, after k-1 substitutions, have $ g = h^{-2k}gh^{2k}$ and not $ g = h^{-2k}g^{-1}h^{2k}$ ?
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 21:43








          • 1




            $begingroup$
            You get $g = h^{-2k}gh^{2k}$ after $2k $ substitutions, not after $ k$, and the reason you get $ g $ and not $ g^{-1}$ is because $ 2k $ is even and after an even number of substitutions you get g while after an odd number of substitutions you get $g^{-1}$
            $endgroup$
            – Sorin Tirc
            Dec 12 '18 at 23:19












          • $begingroup$
            I meant 2k-1 (not k-1) but i was still false....thank you for your help! I got it know.
            $endgroup$
            – FabianSchneider
            Dec 12 '18 at 23:24










          • $begingroup$
            We can´t follow $h^{-1}g^{-1}h=g$ from $h^{-1}gh = g^{-1}$, because G is not neccesarry commutative ?!
            $endgroup$
            – FabianSchneider
            Dec 13 '18 at 16:35








          • 1




            $begingroup$
            you are partly right because it should be $hg^{-1}h^{-1}=g$ but the argument still goes through ;)
            $endgroup$
            – Sorin Tirc
            Dec 13 '18 at 20:39


















          $begingroup$
          Why do we, after k-1 substitutions, have $ g = h^{-2k}gh^{2k}$ and not $ g = h^{-2k}g^{-1}h^{2k}$ ?
          $endgroup$
          – FabianSchneider
          Dec 12 '18 at 21:43






          $begingroup$
          Why do we, after k-1 substitutions, have $ g = h^{-2k}gh^{2k}$ and not $ g = h^{-2k}g^{-1}h^{2k}$ ?
          $endgroup$
          – FabianSchneider
          Dec 12 '18 at 21:43






          1




          1




          $begingroup$
          You get $g = h^{-2k}gh^{2k}$ after $2k $ substitutions, not after $ k$, and the reason you get $ g $ and not $ g^{-1}$ is because $ 2k $ is even and after an even number of substitutions you get g while after an odd number of substitutions you get $g^{-1}$
          $endgroup$
          – Sorin Tirc
          Dec 12 '18 at 23:19






          $begingroup$
          You get $g = h^{-2k}gh^{2k}$ after $2k $ substitutions, not after $ k$, and the reason you get $ g $ and not $ g^{-1}$ is because $ 2k $ is even and after an even number of substitutions you get g while after an odd number of substitutions you get $g^{-1}$
          $endgroup$
          – Sorin Tirc
          Dec 12 '18 at 23:19














          $begingroup$
          I meant 2k-1 (not k-1) but i was still false....thank you for your help! I got it know.
          $endgroup$
          – FabianSchneider
          Dec 12 '18 at 23:24




          $begingroup$
          I meant 2k-1 (not k-1) but i was still false....thank you for your help! I got it know.
          $endgroup$
          – FabianSchneider
          Dec 12 '18 at 23:24












          $begingroup$
          We can´t follow $h^{-1}g^{-1}h=g$ from $h^{-1}gh = g^{-1}$, because G is not neccesarry commutative ?!
          $endgroup$
          – FabianSchneider
          Dec 13 '18 at 16:35






          $begingroup$
          We can´t follow $h^{-1}g^{-1}h=g$ from $h^{-1}gh = g^{-1}$, because G is not neccesarry commutative ?!
          $endgroup$
          – FabianSchneider
          Dec 13 '18 at 16:35






          1




          1




          $begingroup$
          you are partly right because it should be $hg^{-1}h^{-1}=g$ but the argument still goes through ;)
          $endgroup$
          – Sorin Tirc
          Dec 13 '18 at 20:39






          $begingroup$
          you are partly right because it should be $hg^{-1}h^{-1}=g$ but the argument still goes through ;)
          $endgroup$
          – Sorin Tirc
          Dec 13 '18 at 20:39




















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          Quarter-circle Tiles

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          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
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          build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

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          0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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