Group of order 275 act on set of size 18, what is the minimum number of orbit of lenght 1?

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Let $G$ be a group of order 275 acting on set of size 18, what is the minimum number of orbit of length 1?

I think it is 2 because we then have $1+1+5+11$ all of the numbers in the sum are divisors of by $275=5^2*11$.

edited Nov 16 at 10:27

ahulpke

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asked Nov 16 at 10:24

mathnoob

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Let $G$ be a group of order 275 acting on set of size 18, what is the minimum number of orbit of length 1?

I think it is 2 because we then have $1+1+5+11$ all of the numbers in the sum are divisors of by $275=5^2*11$.

edited Nov 16 at 10:27

ahulpke

6,848926

asked Nov 16 at 10:24

mathnoob

70911

New contributor

add a comment |

up vote
0
down vote

favorite

Let $G$ be a group of order 275 acting on set of size 18, what is the minimum number of orbit of length 1?

I think it is 2 because we then have $1+1+5+11$ all of the numbers in the sum are divisors of by $275=5^2*11$.

edited Nov 16 at 10:27

ahulpke

6,848926

asked Nov 16 at 10:24

mathnoob

70911

New contributor

Let $G$ be a group of order 275 acting on set of size 18, what is the minimum number of orbit of length 1?

I think it is 2 because we then have $1+1+5+11$ all of the numbers in the sum are divisors of by $275=5^2*11$.

group-theory finite-groups

edited Nov 16 at 10:27

ahulpke

6,848926

asked Nov 16 at 10:24

mathnoob

70911

New contributor

edited Nov 16 at 10:27

ahulpke

6,848926

asked Nov 16 at 10:24

mathnoob

70911

New contributor

edited Nov 16 at 10:27

ahulpke

6,848926

edited Nov 16 at 10:27

ahulpke

6,848926

edited Nov 16 at 10:27

ahulpke

6,848926

asked Nov 16 at 10:24

mathnoob

70911

New contributor

asked Nov 16 at 10:24

mathnoob

70911

asked Nov 16 at 10:24

mathnoob

70911

New contributor

mathnoob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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You have shown that there is a solution which involves 2 orbits of length one, but not yet that you cannot do with fewer.

For that, note that you cannot write $17=18-1$ as a sum of 5s and 11s (clearly no divisor $>18$ will be relevant and 18 cannot be written as a sum of 5s and 11s).

edited Nov 16 at 14:23

answered Nov 16 at 10:29

ahulpke

6,848926

You also need to check that you cannot write 18 as a sum of 5s and 11s.
– C Monsour
Nov 16 at 11:12

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By orbit stabilizer theorem $|G/Stab(a)|=|O(a)|$
With this we know that orbit must divide f order of group So only possibility is 1 . So there are 18 orbits

answered Nov 16 at 14:56

Shubham

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2 Answers
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2 Answers
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You have shown that there is a solution which involves 2 orbits of length one, but not yet that you cannot do with fewer.

For that, note that you cannot write $17=18-1$ as a sum of 5s and 11s (clearly no divisor $>18$ will be relevant and 18 cannot be written as a sum of 5s and 11s).

edited Nov 16 at 14:23

answered Nov 16 at 10:29

ahulpke

6,848926

You also need to check that you cannot write 18 as a sum of 5s and 11s.
– C Monsour
Nov 16 at 11:12

add a comment |

up vote
0
down vote

You have shown that there is a solution which involves 2 orbits of length one, but not yet that you cannot do with fewer.

For that, note that you cannot write $17=18-1$ as a sum of 5s and 11s (clearly no divisor $>18$ will be relevant and 18 cannot be written as a sum of 5s and 11s).

edited Nov 16 at 14:23

answered Nov 16 at 10:29

ahulpke

6,848926

You also need to check that you cannot write 18 as a sum of 5s and 11s.
– C Monsour
Nov 16 at 11:12

add a comment |

up vote
0
down vote

You have shown that there is a solution which involves 2 orbits of length one, but not yet that you cannot do with fewer.

For that, note that you cannot write $17=18-1$ as a sum of 5s and 11s (clearly no divisor $>18$ will be relevant and 18 cannot be written as a sum of 5s and 11s).

edited Nov 16 at 14:23

answered Nov 16 at 10:29

ahulpke

6,848926

You have shown that there is a solution which involves 2 orbits of length one, but not yet that you cannot do with fewer.

For that, note that you cannot write $17=18-1$ as a sum of 5s and 11s (clearly no divisor $>18$ will be relevant and 18 cannot be written as a sum of 5s and 11s).

edited Nov 16 at 14:23

answered Nov 16 at 10:29

ahulpke

6,848926

edited Nov 16 at 14:23

answered Nov 16 at 10:29

ahulpke

6,848926

answered Nov 16 at 10:29

ahulpke

6,848926

answered Nov 16 at 10:29

ahulpke

6,848926

You also need to check that you cannot write 18 as a sum of 5s and 11s.
– C Monsour
Nov 16 at 11:12

add a comment |

You also need to check that you cannot write 18 as a sum of 5s and 11s.
– C Monsour
Nov 16 at 11:12

You also need to check that you cannot write 18 as a sum of 5s and 11s.
– C Monsour
Nov 16 at 11:12

add a comment |

up vote
0
down vote

By orbit stabilizer theorem $|G/Stab(a)|=|O(a)|$
With this we know that orbit must divide f order of group So only possibility is 1 . So there are 18 orbits

answered Nov 16 at 14:56

Shubham

1,2951518

add a comment |

up vote
0
down vote

By orbit stabilizer theorem $|G/Stab(a)|=|O(a)|$
With this we know that orbit must divide f order of group So only possibility is 1 . So there are 18 orbits

answered Nov 16 at 14:56

Shubham

1,2951518

add a comment |

up vote
0
down vote

By orbit stabilizer theorem $|G/Stab(a)|=|O(a)|$
With this we know that orbit must divide f order of group So only possibility is 1 . So there are 18 orbits

answered Nov 16 at 14:56

Shubham

1,2951518

By orbit stabilizer theorem $|G/Stab(a)|=|O(a)|$
With this we know that orbit must divide f order of group So only possibility is 1 . So there are 18 orbits

answered Nov 16 at 14:56

Shubham

1,2951518

answered Nov 16 at 14:56

Shubham

1,2951518

answered Nov 16 at 14:56

Shubham

1,2951518

answered Nov 16 at 14:56

Shubham

1,2951518

add a comment |

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