If $|G|<infty$ acts transitively on a set $X$ with $|X|=10$ then $G$ has an element of order $5$
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
edited Jan 5 at 21:30
Matt Samuel
37.4k63665
asked Jan 5 at 21:23
John CataldoJohn Cataldo
1,0961216
add a comment |
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
edited Jan 5 at 21:30
Matt Samuel
37.4k63665
asked Jan 5 at 21:23
John CataldoJohn Cataldo
1,0961216
add a comment |
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
edited Jan 5 at 21:30
Matt Samuel
37.4k63665
asked Jan 5 at 21:23
John CataldoJohn Cataldo
1,0961216
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
edited Jan 5 at 21:30
Matt Samuel
37.4k63665
asked Jan 5 at 21:23
John CataldoJohn Cataldo
1,0961216
edited Jan 5 at 21:30
Matt Samuel
37.4k63665
asked Jan 5 at 21:23
John CataldoJohn Cataldo
1,0961216
edited Jan 5 at 21:30
Matt Samuel
37.4k63665
edited Jan 5 at 21:30
Matt Samuel
37.4k63665
edited Jan 5 at 21:30
Matt Samuel
37.4k63665
37.4k63665
asked Jan 5 at 21:23
John CataldoJohn Cataldo
1,0961216
asked Jan 5 at 21:23
John CataldoJohn Cataldo
1,0961216
asked Jan 5 at 21:23
John CataldoJohn Cataldo
1,0961216
1,0961216
add a comment |
add a comment |
1 Answer
1
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oldest
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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 $5$.
answered Jan 5 at 21:26
Matt SamuelMatt Samuel
37.4k63665
Could you have used any of the Sylow theorems for this?
– John Cataldo
2 days ago
@John Have you not encountered the orbit stabilizer or Cauchy's theorem?
– Matt Samuel
2 days ago
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
– John Cataldo
2 days ago
@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.
– Matt Samuel
2 days ago
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
votes
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 $5$.
answered Jan 5 at 21:26
Matt SamuelMatt Samuel
37.4k63665
Could you have used any of the Sylow theorems for this?
– John Cataldo
2 days ago
@John Have you not encountered the orbit stabilizer or Cauchy's theorem?
– Matt Samuel
2 days ago
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
– John Cataldo
2 days ago
@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.
– Matt Samuel
2 days ago
add a comment |
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 $5$.
answered Jan 5 at 21:26
Matt SamuelMatt Samuel
37.4k63665
Could you have used any of the Sylow theorems for this?
– John Cataldo
2 days ago
@John Have you not encountered the orbit stabilizer or Cauchy's theorem?
– Matt Samuel
2 days ago
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
– John Cataldo
2 days ago
@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.
– Matt Samuel
2 days ago
add a comment |
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 $5$.
answered Jan 5 at 21:26
Matt SamuelMatt Samuel
37.4k63665
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 $5$.
answered Jan 5 at 21:26
Matt SamuelMatt Samuel
37.4k63665
answered Jan 5 at 21:26
Matt SamuelMatt Samuel
37.4k63665
answered Jan 5 at 21:26
Matt SamuelMatt Samuel
37.4k63665
answered Jan 5 at 21:26
Matt SamuelMatt Samuel
37.4k63665
37.4k63665
Could you have used any of the Sylow theorems for this?
– John Cataldo
2 days ago
@John Have you not encountered the orbit stabilizer or Cauchy's theorem?
– Matt Samuel
2 days ago
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
– John Cataldo
2 days ago
@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.
– Matt Samuel
2 days ago
add a comment |
Could you have used any of the Sylow theorems for this?
– John Cataldo
2 days ago
@John Have you not encountered the orbit stabilizer or Cauchy's theorem?
– Matt Samuel
2 days ago
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
– John Cataldo
2 days ago
@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.
– Matt Samuel
2 days ago
Could you have used any of the Sylow theorems for this?
– John Cataldo
2 days ago
Could you have used any of the Sylow theorems for this?
– John Cataldo
2 days ago
@John Have you not encountered the orbit stabilizer or Cauchy's theorem?
– Matt Samuel
2 days ago
@John Have you not encountered the orbit stabilizer or Cauchy's theorem?
– Matt Samuel
2 days ago
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
– John Cataldo
2 days ago
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
– John Cataldo
2 days ago
@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.
– Matt Samuel
2 days ago
@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.
– Matt Samuel
2 days ago
add a comment |
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