If $|G|<infty$ acts transitively on a set $X$ with $|X|=10$ then $G$ has an element of order $5$












3












$begingroup$



Let $G$ be a finite group that acts transitively on a set $X$ with $|X|=10$. Show that $exists gin G$ of order $5$




There is a homomorphism $phi:Gmapsto S_{10}$ that sends $gin G$ to its corresponding permutation in $S_{10}$.



So $G$ is isomorphic to a subgroup of $S_{10}$. I want to prove that $G$ contains a $5$-cycle.



Since $G$ acts transitively $forall x,yin {1,2,...,10}~exists gin G$ such that $gx=y.$ Let $x$ be in ${1,2,...,10}$ Let's consider the set $langle grangle x ={x,gx,g^2x,...}$. If it contains $X$ then $g$ is a $10$-cycle and $g^2$ has order $5$.



There must be some cycles in $G$ of length $ge 3$ because if we only have disjoint transpositions $G$ doesn't act transitively.



I think I need some kind of hint.






abstract-algebra group-theory symmetric-groups group-actions







share|cite|improve this question











$endgroup$

















    3












    $begingroup$



    Let $G$ be a finite group that acts transitively on a set $X$ with $|X|=10$. Show that $exists gin G$ of order $5$




    There is a homomorphism $phi:Gmapsto S_{10}$ that sends $gin G$ to its corresponding permutation in $S_{10}$.



    So $G$ is isomorphic to a subgroup of $S_{10}$. I want to prove that $G$ contains a $5$-cycle.



    Since $G$ acts transitively $forall x,yin {1,2,...,10}~exists gin G$ such that $gx=y.$ Let $x$ be in ${1,2,...,10}$ Let's consider the set $langle grangle x ={x,gx,g^2x,...}$. If it contains $X$ then $g$ is a $10$-cycle and $g^2$ has order $5$.



    There must be some cycles in $G$ of length $ge 3$ because if we only have disjoint transpositions $G$ doesn't act transitively.



    I think I need some kind of hint.






    abstract-algebra group-theory symmetric-groups group-actions







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      Let $G$ be a finite group that acts transitively on a set $X$ with $|X|=10$. Show that $exists gin G$ of order $5$




      There is a homomorphism $phi:Gmapsto S_{10}$ that sends $gin G$ to its corresponding permutation in $S_{10}$.



      So $G$ is isomorphic to a subgroup of $S_{10}$. I want to prove that $G$ contains a $5$-cycle.



      Since $G$ acts transitively $forall x,yin {1,2,...,10}~exists gin G$ such that $gx=y.$ Let $x$ be in ${1,2,...,10}$ Let's consider the set $langle grangle x ={x,gx,g^2x,...}$. If it contains $X$ then $g$ is a $10$-cycle and $g^2$ has order $5$.



      There must be some cycles in $G$ of length $ge 3$ because if we only have disjoint transpositions $G$ doesn't act transitively.



      I think I need some kind of hint.






      abstract-algebra group-theory symmetric-groups group-actions







      share|cite|improve this question











      $endgroup$





      Let $G$ be a finite group that acts transitively on a set $X$ with $|X|=10$. Show that $exists gin G$ of order $5$




      There is a homomorphism $phi:Gmapsto S_{10}$ that sends $gin G$ to its corresponding permutation in $S_{10}$.



      So $G$ is isomorphic to a subgroup of $S_{10}$. I want to prove that $G$ contains a $5$-cycle.



      Since $G$ acts transitively $forall x,yin {1,2,...,10}~exists gin G$ such that $gx=y.$ Let $x$ be in ${1,2,...,10}$ Let's consider the set $langle grangle x ={x,gx,g^2x,...}$. If it contains $X$ then $g$ is a $10$-cycle and $g^2$ has order $5$.



      There must be some cycles in $G$ of length $ge 3$ because if we only have disjoint transpositions $G$ doesn't act transitively.



      I think I need some kind of hint.






      abstract-algebra group-theory symmetric-groups group-actions




      abstract-algebra group-theory symmetric-groups group-actions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 21:30









      Matt Samuel

      38.8k63769




      38.8k63769










      asked Jan 5 at 21:23









      John CataldoJohn Cataldo

      1,1881316




      1,1881316






















          1 Answer
          1






          active

          oldest

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          7












          $begingroup$

          This is actually easy. By the orbit-stabilizer theorem $G$ has order divisible by $10$. By Cauchy's theorem it therefore has an element of order $5$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you have used any of the Sylow theorems for this?
            $endgroup$
            – John Cataldo
            Jan 6 at 7:20










          • $begingroup$
            @John Have you not encountered the orbit stabilizer or Cauchy's theorem?
            $endgroup$
            – Matt Samuel
            Jan 6 at 9:41










          • $begingroup$
            Yes I have, well, I read about the Cauchy theorem outside of class. But we did prove the Sylow theorems and the orbit-stabilizer theorem so I was interested in a proof using those. And it seems like Cauchy and Sylow are very similar
            $endgroup$
            – John Cataldo
            Jan 6 at 9:43












          • $begingroup$
            @John The strong version of Sylow's first theorem says that there is a subgroup of order $5$, and that is necessarily cyclic. That usually comes up in the proof. If you just use them out of the box, though, Sylow's theorems don't prove it. Cauchy's theorem is very basic and usually proved before Sylow's theorems.
            $endgroup$
            – Matt Samuel
            Jan 6 at 11:40











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          This is actually easy. By the orbit-stabilizer theorem $G$ has order divisible by $10$. By Cauchy's theorem it therefore has an element of order $5$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you have used any of the Sylow theorems for this?
            $endgroup$
            – John Cataldo
            Jan 6 at 7:20










          • $begingroup$
            @John Have you not encountered the orbit stabilizer or Cauchy's theorem?
            $endgroup$
            – Matt Samuel
            Jan 6 at 9:41










          • $begingroup$
            Yes I have, well, I read about the Cauchy theorem outside of class. But we did prove the Sylow theorems and the orbit-stabilizer theorem so I was interested in a proof using those. And it seems like Cauchy and Sylow are very similar
            $endgroup$
            – John Cataldo
            Jan 6 at 9:43












          • $begingroup$
            @John The strong version of Sylow's first theorem says that there is a subgroup of order $5$, and that is necessarily cyclic. That usually comes up in the proof. If you just use them out of the box, though, Sylow's theorems don't prove it. Cauchy's theorem is very basic and usually proved before Sylow's theorems.
            $endgroup$
            – Matt Samuel
            Jan 6 at 11:40
















          7












          $begingroup$

          This is actually easy. By the orbit-stabilizer theorem $G$ has order divisible by $10$. By Cauchy's theorem it therefore has an element of order $5$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you have used any of the Sylow theorems for this?
            $endgroup$
            – John Cataldo
            Jan 6 at 7:20










          • $begingroup$
            @John Have you not encountered the orbit stabilizer or Cauchy's theorem?
            $endgroup$
            – Matt Samuel
            Jan 6 at 9:41










          • $begingroup$
            Yes I have, well, I read about the Cauchy theorem outside of class. But we did prove the Sylow theorems and the orbit-stabilizer theorem so I was interested in a proof using those. And it seems like Cauchy and Sylow are very similar
            $endgroup$
            – John Cataldo
            Jan 6 at 9:43












          • $begingroup$
            @John The strong version of Sylow's first theorem says that there is a subgroup of order $5$, and that is necessarily cyclic. That usually comes up in the proof. If you just use them out of the box, though, Sylow's theorems don't prove it. Cauchy's theorem is very basic and usually proved before Sylow's theorems.
            $endgroup$
            – Matt Samuel
            Jan 6 at 11:40














          7












          7








          7





          $begingroup$

          This is actually easy. By the orbit-stabilizer theorem $G$ has order divisible by $10$. By Cauchy's theorem it therefore has an element of order $5$.






          share|cite|improve this answer









          $endgroup$



          This is actually easy. By the orbit-stabilizer theorem $G$ has order divisible by $10$. By Cauchy's theorem it therefore has an element of order $5$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 at 21:26









          Matt SamuelMatt Samuel

          38.8k63769




          38.8k63769












          • $begingroup$
            Could you have used any of the Sylow theorems for this?
            $endgroup$
            – John Cataldo
            Jan 6 at 7:20










          • $begingroup$
            @John Have you not encountered the orbit stabilizer or Cauchy's theorem?
            $endgroup$
            – Matt Samuel
            Jan 6 at 9:41










          • $begingroup$
            Yes I have, well, I read about the Cauchy theorem outside of class. But we did prove the Sylow theorems and the orbit-stabilizer theorem so I was interested in a proof using those. And it seems like Cauchy and Sylow are very similar
            $endgroup$
            – John Cataldo
            Jan 6 at 9:43












          • $begingroup$
            @John The strong version of Sylow's first theorem says that there is a subgroup of order $5$, and that is necessarily cyclic. That usually comes up in the proof. If you just use them out of the box, though, Sylow's theorems don't prove it. Cauchy's theorem is very basic and usually proved before Sylow's theorems.
            $endgroup$
            – Matt Samuel
            Jan 6 at 11:40


















          • $begingroup$
            Could you have used any of the Sylow theorems for this?
            $endgroup$
            – John Cataldo
            Jan 6 at 7:20










          • $begingroup$
            @John Have you not encountered the orbit stabilizer or Cauchy's theorem?
            $endgroup$
            – Matt Samuel
            Jan 6 at 9:41










          • $begingroup$
            Yes I have, well, I read about the Cauchy theorem outside of class. But we did prove the Sylow theorems and the orbit-stabilizer theorem so I was interested in a proof using those. And it seems like Cauchy and Sylow are very similar
            $endgroup$
            – John Cataldo
            Jan 6 at 9:43












          • $begingroup$
            @John The strong version of Sylow's first theorem says that there is a subgroup of order $5$, and that is necessarily cyclic. That usually comes up in the proof. If you just use them out of the box, though, Sylow's theorems don't prove it. Cauchy's theorem is very basic and usually proved before Sylow's theorems.
            $endgroup$
            – Matt Samuel
            Jan 6 at 11:40
















          $begingroup$
          Could you have used any of the Sylow theorems for this?
          $endgroup$
          – John Cataldo
          Jan 6 at 7:20




          $begingroup$
          Could you have used any of the Sylow theorems for this?
          $endgroup$
          – John Cataldo
          Jan 6 at 7:20












          $begingroup$
          @John Have you not encountered the orbit stabilizer or Cauchy's theorem?
          $endgroup$
          – Matt Samuel
          Jan 6 at 9:41




          $begingroup$
          @John Have you not encountered the orbit stabilizer or Cauchy's theorem?
          $endgroup$
          – Matt Samuel
          Jan 6 at 9:41












          $begingroup$
          Yes I have, well, I read about the Cauchy theorem outside of class. But we did prove the Sylow theorems and the orbit-stabilizer theorem so I was interested in a proof using those. And it seems like Cauchy and Sylow are very similar
          $endgroup$
          – John Cataldo
          Jan 6 at 9:43






          $begingroup$
          Yes I have, well, I read about the Cauchy theorem outside of class. But we did prove the Sylow theorems and the orbit-stabilizer theorem so I was interested in a proof using those. And it seems like Cauchy and Sylow are very similar
          $endgroup$
          – John Cataldo
          Jan 6 at 9:43














          $begingroup$
          @John The strong version of Sylow's first theorem says that there is a subgroup of order $5$, and that is necessarily cyclic. That usually comes up in the proof. If you just use them out of the box, though, Sylow's theorems don't prove it. Cauchy's theorem is very basic and usually proved before Sylow's theorems.
          $endgroup$
          – Matt Samuel
          Jan 6 at 11:40




          $begingroup$
          @John The strong version of Sylow's first theorem says that there is a subgroup of order $5$, and that is necessarily cyclic. That usually comes up in the proof. If you just use them out of the box, though, Sylow's theorems don't prove it. Cauchy's theorem is very basic and usually proved before Sylow's theorems.
          $endgroup$
          – Matt Samuel
          Jan 6 at 11:40


















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