Relationship between order of a group acting on a set and faithfulness
I am trying to prove (or to disprove) that $forall n,m in mathbb{N}$ where $gcd (n, m) = 1$ and $n, m geq 2$, $exists G$ a group such that $|G| = n$ and $exists X$ a set such that $|X| = m$ such that $G$ acts on $X$ such that there are no trivial orbits (no fixed points). I do not know where to start.
If this is untrue, then is there some restriction such that it becomes true?
group-theory finite-groups
asked Nov 18 at 23:12
Kiarash Jamali
334
add a comment |
I am trying to prove (or to disprove) that $forall n,m in mathbb{N}$ where $gcd (n, m) = 1$ and $n, m geq 2$, $exists G$ a group such that $|G| = n$ and $exists X$ a set such that $|X| = m$ such that $G$ acts on $X$ such that there are no trivial orbits (no fixed points). I do not know where to start.
If this is untrue, then is there some restriction such that it becomes true?
group-theory finite-groups
asked Nov 18 at 23:12
Kiarash Jamali
334
add a comment |
I am trying to prove (or to disprove) that $forall n,m in mathbb{N}$ where $gcd (n, m) = 1$ and $n, m geq 2$, $exists G$ a group such that $|G| = n$ and $exists X$ a set such that $|X| = m$ such that $G$ acts on $X$ such that there are no trivial orbits (no fixed points). I do not know where to start.
If this is untrue, then is there some restriction such that it becomes true?
group-theory finite-groups
asked Nov 18 at 23:12
Kiarash Jamali
334
I am trying to prove (or to disprove) that $forall n,m in mathbb{N}$ where $gcd (n, m) = 1$ and $n, m geq 2$, $exists G$ a group such that $|G| = n$ and $exists X$ a set such that $|X| = m$ such that $G$ acts on $X$ such that there are no trivial orbits (no fixed points). I do not know where to start.
If this is untrue, then is there some restriction such that it becomes true?
group-theory finite-groups
group-theory finite-groups
asked Nov 18 at 23:12
Kiarash Jamali
334
asked Nov 18 at 23:12
Kiarash Jamali
334
edited Nov 19 at 2:20
asked Nov 18 at 23:12
Kiarash Jamali
334
asked Nov 18 at 23:12
Kiarash Jamali
334
asked Nov 18 at 23:12
Kiarash Jamali
334
334
add a comment |
add a comment |
2 Answers
2
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I $G$ acts faithfully on a set $X$ then we can define a new action on any set $Xsubset Y$ setting $g(a)=a$ for any $ain Ysetminus X$ and $Y$ can have any cardinality
answered Nov 18 at 23:23
ALG
55413
add a comment |
Let $G=S_3$ and let $X=S_3/langle (12)ranglecup S_3/langle (123)rangle$. This set has five elements and the action is faithful.
answered Nov 18 at 23:24
Matt Samuel
37.1k63465
So would this be true for every group and set that have orders that are co-prime?
– Kiarash Jamali
Nov 19 at 2:03
@Kiar No, just one group and one set. You use $exists$ in your question, which implies that only one example is required.
– Matt Samuel
Nov 19 at 2:04
@Kia And the question has changed since I answered it.
– Matt Samuel
Nov 19 at 2:05
Sorry, I rephrased it. I'm trying to see if there is some sort of relationship between the order of the group and set and it being faithful or not
– Kiarash Jamali
Nov 19 at 2:05
@Kia As the other answer says, the divisibility had nothing to do with it. Only the size.
– Matt Samuel
Nov 19 at 2:06
|
show 6 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I $G$ acts faithfully on a set $X$ then we can define a new action on any set $Xsubset Y$ setting $g(a)=a$ for any $ain Ysetminus X$ and $Y$ can have any cardinality
answered Nov 18 at 23:23
ALG
55413
add a comment |
I $G$ acts faithfully on a set $X$ then we can define a new action on any set $Xsubset Y$ setting $g(a)=a$ for any $ain Ysetminus X$ and $Y$ can have any cardinality
answered Nov 18 at 23:23
ALG
55413
add a comment |
I $G$ acts faithfully on a set $X$ then we can define a new action on any set $Xsubset Y$ setting $g(a)=a$ for any $ain Ysetminus X$ and $Y$ can have any cardinality
answered Nov 18 at 23:23
ALG
55413
I $G$ acts faithfully on a set $X$ then we can define a new action on any set $Xsubset Y$ setting $g(a)=a$ for any $ain Ysetminus X$ and $Y$ can have any cardinality
answered Nov 18 at 23:23
ALG
55413
answered Nov 18 at 23:23
ALG
55413
answered Nov 18 at 23:23
ALG
55413
answered Nov 18 at 23:23
ALG
55413
55413
add a comment |
add a comment |
Let $G=S_3$ and let $X=S_3/langle (12)ranglecup S_3/langle (123)rangle$. This set has five elements and the action is faithful.
answered Nov 18 at 23:24
Matt Samuel
37.1k63465
So would this be true for every group and set that have orders that are co-prime?
– Kiarash Jamali
Nov 19 at 2:03
@Kiar No, just one group and one set. You use $exists$ in your question, which implies that only one example is required.
– Matt Samuel
Nov 19 at 2:04
@Kia And the question has changed since I answered it.
– Matt Samuel
Nov 19 at 2:05
Sorry, I rephrased it. I'm trying to see if there is some sort of relationship between the order of the group and set and it being faithful or not
– Kiarash Jamali
Nov 19 at 2:05
@Kia As the other answer says, the divisibility had nothing to do with it. Only the size.
– Matt Samuel
Nov 19 at 2:06
|
show 6 more comments
Let $G=S_3$ and let $X=S_3/langle (12)ranglecup S_3/langle (123)rangle$. This set has five elements and the action is faithful.
answered Nov 18 at 23:24
Matt Samuel
37.1k63465
So would this be true for every group and set that have orders that are co-prime?
– Kiarash Jamali
Nov 19 at 2:03
@Kiar No, just one group and one set. You use $exists$ in your question, which implies that only one example is required.
– Matt Samuel
Nov 19 at 2:04
@Kia And the question has changed since I answered it.
– Matt Samuel
Nov 19 at 2:05
Sorry, I rephrased it. I'm trying to see if there is some sort of relationship between the order of the group and set and it being faithful or not
– Kiarash Jamali
Nov 19 at 2:05
@Kia As the other answer says, the divisibility had nothing to do with it. Only the size.
– Matt Samuel
Nov 19 at 2:06
|
show 6 more comments
Let $G=S_3$ and let $X=S_3/langle (12)ranglecup S_3/langle (123)rangle$. This set has five elements and the action is faithful.
answered Nov 18 at 23:24
Matt Samuel
37.1k63465
Let $G=S_3$ and let $X=S_3/langle (12)ranglecup S_3/langle (123)rangle$. This set has five elements and the action is faithful.
answered Nov 18 at 23:24
Matt Samuel
37.1k63465
answered Nov 18 at 23:24
Matt Samuel
37.1k63465
answered Nov 18 at 23:24
Matt Samuel
37.1k63465
answered Nov 18 at 23:24
Matt Samuel
37.1k63465
37.1k63465
So would this be true for every group and set that have orders that are co-prime?
– Kiarash Jamali
Nov 19 at 2:03
@Kiar No, just one group and one set. You use $exists$ in your question, which implies that only one example is required.
– Matt Samuel
Nov 19 at 2:04
@Kia And the question has changed since I answered it.
– Matt Samuel
Nov 19 at 2:05
Sorry, I rephrased it. I'm trying to see if there is some sort of relationship between the order of the group and set and it being faithful or not
– Kiarash Jamali
Nov 19 at 2:05
@Kia As the other answer says, the divisibility had nothing to do with it. Only the size.
– Matt Samuel
Nov 19 at 2:06
|
show 6 more comments
So would this be true for every group and set that have orders that are co-prime?
– Kiarash Jamali
Nov 19 at 2:03
@Kiar No, just one group and one set. You use $exists$ in your question, which implies that only one example is required.
– Matt Samuel
Nov 19 at 2:04
@Kia And the question has changed since I answered it.
– Matt Samuel
Nov 19 at 2:05
Sorry, I rephrased it. I'm trying to see if there is some sort of relationship between the order of the group and set and it being faithful or not
– Kiarash Jamali
Nov 19 at 2:05
@Kia As the other answer says, the divisibility had nothing to do with it. Only the size.
– Matt Samuel
Nov 19 at 2:06
So would this be true for every group and set that have orders that are co-prime?
– Kiarash Jamali
Nov 19 at 2:03
So would this be true for every group and set that have orders that are co-prime?
– Kiarash Jamali
Nov 19 at 2:03
@Kiar No, just one group and one set. You use $exists$ in your question, which implies that only one example is required.
– Matt Samuel
Nov 19 at 2:04
@Kiar No, just one group and one set. You use $exists$ in your question, which implies that only one example is required.
– Matt Samuel
Nov 19 at 2:04
@Kia And the question has changed since I answered it.
– Matt Samuel
Nov 19 at 2:05
@Kia And the question has changed since I answered it.
– Matt Samuel
Nov 19 at 2:05
Sorry, I rephrased it. I'm trying to see if there is some sort of relationship between the order of the group and set and it being faithful or not
– Kiarash Jamali
Nov 19 at 2:05
Sorry, I rephrased it. I'm trying to see if there is some sort of relationship between the order of the group and set and it being faithful or not
– Kiarash Jamali
Nov 19 at 2:05
@Kia As the other answer says, the divisibility had nothing to do with it. Only the size.
– Matt Samuel
Nov 19 at 2:06
@Kia As the other answer says, the divisibility had nothing to do with it. Only the size.
– Matt Samuel
Nov 19 at 2:06
|
show 6 more comments
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