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Proving a mapping is a group action

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4

$begingroup$

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

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asked 4 hours ago

cb7cb7

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$endgroup$

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4

$begingroup$

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

share|cite|improve this question

asked 4 hours ago

cb7cb7

1256

$endgroup$

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

share|cite|improve this question

asked 4 hours ago

cb7cb7

1256

$endgroup$

share|cite|improve this question

asked 4 hours ago

cb7cb7

1256

share|cite|improve this question

asked 4 hours ago

cb7cb7

1256

asked 4 hours ago

cb7cb7

1256

asked 4 hours ago

cb7cb7

1256

1 Answer
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5

$begingroup$

Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.

share|cite|improve this answer

answered 4 hours ago

MarkMark

9,579622

$endgroup$

add a comment | 

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5

$begingroup$

Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.

share|cite|improve this answer

answered 4 hours ago

MarkMark

9,579622

$endgroup$

add a comment | 

5

$begingroup$

Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.

share|cite|improve this answer

answered 4 hours ago

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 $t_1=t_2$.

share|cite|improve this answer

answered 4 hours ago

MarkMark

9,579622

$endgroup$

share|cite|improve this answer

answered 4 hours ago

MarkMark

9,579622

answered 4 hours ago

MarkMark

9,579622

answered 4 hours ago

MarkMark

9,579622