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Proving a mapping is a group action
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$begingroup$
Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.
My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.
For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.
Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?
group-theory finite-groups group-actions
asked yesterday
cb7cb7
1306
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.
My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.
For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.
Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?
group-theory finite-groups group-actions
asked yesterday
cb7cb7
1306
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.
My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.
For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.
Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?
group-theory finite-groups group-actions
asked yesterday
cb7cb7
1306
$endgroup$
Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.
My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.
For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.
Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?
group-theory finite-groups group-actions
group-theory finite-groups group-actions
asked yesterday
cb7cb7
1306
asked yesterday
cb7cb7
1306
asked yesterday
cb7cb7
1306
asked yesterday
cb7cb7
1306
asked yesterday
cb7cb7
1306
1306
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.
answered yesterday
MarkMark
9,579622
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.
answered yesterday
MarkMark
9,579622
$endgroup$
add a comment |
$begingroup$
Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.
answered yesterday
MarkMark
9,579622
$endgroup$
add a comment |
$begingroup$
Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.
answered yesterday
MarkMark
9,579622
$endgroup$
Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.
answered yesterday
MarkMark
9,579622
answered yesterday
MarkMark
9,579622
answered yesterday
MarkMark
9,579622
answered yesterday
MarkMark
9,579622
9,579622
add a comment |
add a comment |
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