Can a group act on the empty set?
$begingroup$
There isn't much more to add to this question. Can we define an action between some group and the null set?
I would have thought that there being no elements to act on it trivially satisfies the requirements for something to be an action but I'm not sure.
group-theory group-actions
edited Mar 10 at 10:50
rabota
14.2k32782
asked Mar 10 at 10:43
andrewandrew
999
$endgroup$
|
show 2 more comments
$begingroup$
There isn't much more to add to this question. Can we define an action between some group and the null set?
I would have thought that there being no elements to act on it trivially satisfies the requirements for something to be an action but I'm not sure.
group-theory group-actions
edited Mar 10 at 10:50
rabota
14.2k32782
asked Mar 10 at 10:43
andrewandrew
999
$endgroup$
2
$begingroup$
Though it's kind of empty to have a group action on an empty set, isn't it? =)
$endgroup$
– user21820
Mar 10 at 11:30
3
$begingroup$
In particular, the symmetric group $S_0$, which has order $1$, acts naturally on the empty set. There is unique bijection between the empty set and itself.
$endgroup$
– Derek Holt
Mar 10 at 11:57
$begingroup$
@user21820 the interest of a mathematical formalism is to avoid such philosophical considerations. In the same spirit, there were mathematicians fighting against the existence of infinite sets in the late XIX...
$endgroup$
– YCor
Mar 10 at 13:41
$begingroup$
@YCor: Erm... I was just joking in my first comment, but I disagree with your comment, because anyone who claims they use ZFC as their foundational system necessarily has made some very weird philosophical assumptions whether or not they know it.
$endgroup$
– user21820
Mar 10 at 14:10
1
$begingroup$
@YCor: But that's only if you think "truth within set theory" is meaningful. To refrain from prolonging this thread with our off-topic discussion, do you want to come to the logic chat-room?
$endgroup$
– user21820
Mar 10 at 14:16
|
show 2 more comments
$begingroup$
There isn't much more to add to this question. Can we define an action between some group and the null set?
I would have thought that there being no elements to act on it trivially satisfies the requirements for something to be an action but I'm not sure.
group-theory group-actions
edited Mar 10 at 10:50
rabota
14.2k32782
asked Mar 10 at 10:43
andrewandrew
999
$endgroup$
There isn't much more to add to this question. Can we define an action between some group and the null set?
I would have thought that there being no elements to act on it trivially satisfies the requirements for something to be an action but I'm not sure.
group-theory group-actions
group-theory group-actions
edited Mar 10 at 10:50
rabota
14.2k32782
asked Mar 10 at 10:43
andrewandrew
999
edited Mar 10 at 10:50
rabota
14.2k32782
asked Mar 10 at 10:43
andrewandrew
999
edited Mar 10 at 10:50
rabota
14.2k32782
edited Mar 10 at 10:50
rabota
14.2k32782
edited Mar 10 at 10:50
rabota
14.2k32782
14.2k32782
asked Mar 10 at 10:43
andrewandrew
999
asked Mar 10 at 10:43
andrewandrew
999
asked Mar 10 at 10:43
andrewandrew
999
999
2
$begingroup$
Though it's kind of empty to have a group action on an empty set, isn't it? =)
$endgroup$
– user21820
Mar 10 at 11:30
3
$begingroup$
In particular, the symmetric group $S_0$, which has order $1$, acts naturally on the empty set. There is unique bijection between the empty set and itself.
$endgroup$
– Derek Holt
Mar 10 at 11:57
$begingroup$
@user21820 the interest of a mathematical formalism is to avoid such philosophical considerations. In the same spirit, there were mathematicians fighting against the existence of infinite sets in the late XIX...
$endgroup$
– YCor
Mar 10 at 13:41
$begingroup$
@YCor: Erm... I was just joking in my first comment, but I disagree with your comment, because anyone who claims they use ZFC as their foundational system necessarily has made some very weird philosophical assumptions whether or not they know it.
$endgroup$
– user21820
Mar 10 at 14:10
1
$begingroup$
@YCor: But that's only if you think "truth within set theory" is meaningful. To refrain from prolonging this thread with our off-topic discussion, do you want to come to the logic chat-room?
$endgroup$
– user21820
Mar 10 at 14:16
|
show 2 more comments
2
$begingroup$
Though it's kind of empty to have a group action on an empty set, isn't it? =)
$endgroup$
– user21820
Mar 10 at 11:30
3
$begingroup$
In particular, the symmetric group $S_0$, which has order $1$, acts naturally on the empty set. There is unique bijection between the empty set and itself.
$endgroup$
– Derek Holt
Mar 10 at 11:57
$begingroup$
@user21820 the interest of a mathematical formalism is to avoid such philosophical considerations. In the same spirit, there were mathematicians fighting against the existence of infinite sets in the late XIX...
$endgroup$
– YCor
Mar 10 at 13:41
$begingroup$
@YCor: Erm... I was just joking in my first comment, but I disagree with your comment, because anyone who claims they use ZFC as their foundational system necessarily has made some very weird philosophical assumptions whether or not they know it.
$endgroup$
– user21820
Mar 10 at 14:10
1
$begingroup$
@YCor: But that's only if you think "truth within set theory" is meaningful. To refrain from prolonging this thread with our off-topic discussion, do you want to come to the logic chat-room?
$endgroup$
– user21820
Mar 10 at 14:16
2
2
$begingroup$
Though it's kind of empty to have a group action on an empty set, isn't it? =)
$endgroup$
– user21820
Mar 10 at 11:30
$begingroup$
Though it's kind of empty to have a group action on an empty set, isn't it? =)
$endgroup$
– user21820
Mar 10 at 11:30
3
3
$begingroup$
In particular, the symmetric group $S_0$, which has order $1$, acts naturally on the empty set. There is unique bijection between the empty set and itself.
$endgroup$
– Derek Holt
Mar 10 at 11:57
$begingroup$
In particular, the symmetric group $S_0$, which has order $1$, acts naturally on the empty set. There is unique bijection between the empty set and itself.
$endgroup$
– Derek Holt
Mar 10 at 11:57
$begingroup$
@user21820 the interest of a mathematical formalism is to avoid such philosophical considerations. In the same spirit, there were mathematicians fighting against the existence of infinite sets in the late XIX...
$endgroup$
– YCor
Mar 10 at 13:41
$begingroup$
@user21820 the interest of a mathematical formalism is to avoid such philosophical considerations. In the same spirit, there were mathematicians fighting against the existence of infinite sets in the late XIX...
$endgroup$
– YCor
Mar 10 at 13:41
$begingroup$
@YCor: Erm... I was just joking in my first comment, but I disagree with your comment, because anyone who claims they use ZFC as their foundational system necessarily has made some very weird philosophical assumptions whether or not they know it.
$endgroup$
– user21820
Mar 10 at 14:10
$begingroup$
@YCor: Erm... I was just joking in my first comment, but I disagree with your comment, because anyone who claims they use ZFC as their foundational system necessarily has made some very weird philosophical assumptions whether or not they know it.
$endgroup$
– user21820
Mar 10 at 14:10
1
1
$begingroup$
@YCor: But that's only if you think "truth within set theory" is meaningful. To refrain from prolonging this thread with our off-topic discussion, do you want to come to the logic chat-room?
$endgroup$
– user21820
Mar 10 at 14:16
$begingroup$
@YCor: But that's only if you think "truth within set theory" is meaningful. To refrain from prolonging this thread with our off-topic discussion, do you want to come to the logic chat-room?
$endgroup$
– user21820
Mar 10 at 14:16
|
show 2 more comments
1 Answer
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$begingroup$
yes you can define the trivial action.
Note that the axioms for group action begins with "for all"
That is:
For all $xin emptyset$ we have that $e.x=x$.
For all $xinemptyset$ and all $g,hin G$ we have $(gh)x=g.(h.x)$
Both statements hold trivially.
answered Mar 10 at 10:46
YankoYanko
7,8801830
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add a comment |
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$begingroup$
yes you can define the trivial action.
Note that the axioms for group action begins with "for all"
That is:
For all $xin emptyset$ we have that $e.x=x$.
For all $xinemptyset$ and all $g,hin G$ we have $(gh)x=g.(h.x)$
Both statements hold trivially.
answered Mar 10 at 10:46
YankoYanko
7,8801830
$endgroup$
add a comment |
$begingroup$
yes you can define the trivial action.
Note that the axioms for group action begins with "for all"
That is:
For all $xin emptyset$ we have that $e.x=x$.
For all $xinemptyset$ and all $g,hin G$ we have $(gh)x=g.(h.x)$
Both statements hold trivially.
answered Mar 10 at 10:46
YankoYanko
7,8801830
$endgroup$
add a comment |
$begingroup$
yes you can define the trivial action.
Note that the axioms for group action begins with "for all"
That is:
For all $xin emptyset$ we have that $e.x=x$.
For all $xinemptyset$ and all $g,hin G$ we have $(gh)x=g.(h.x)$
Both statements hold trivially.
answered Mar 10 at 10:46
YankoYanko
7,8801830
$endgroup$
yes you can define the trivial action.
Note that the axioms for group action begins with "for all"
That is:
For all $xin emptyset$ we have that $e.x=x$.
For all $xinemptyset$ and all $g,hin G$ we have $(gh)x=g.(h.x)$
Both statements hold trivially.
answered Mar 10 at 10:46
YankoYanko
7,8801830
answered Mar 10 at 10:46
YankoYanko
7,8801830
answered Mar 10 at 10:46
YankoYanko
7,8801830
answered Mar 10 at 10:46
YankoYanko
7,8801830
7,8801830
add a comment |
add a comment |
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$begingroup$
Though it's kind of empty to have a group action on an empty set, isn't it? =)
$endgroup$
– user21820
Mar 10 at 11:30
3
$begingroup$
In particular, the symmetric group $S_0$, which has order $1$, acts naturally on the empty set. There is unique bijection between the empty set and itself.
$endgroup$
– Derek Holt
Mar 10 at 11:57
$begingroup$
@user21820 the interest of a mathematical formalism is to avoid such philosophical considerations. In the same spirit, there were mathematicians fighting against the existence of infinite sets in the late XIX...
$endgroup$
– YCor
Mar 10 at 13:41
$begingroup$
@YCor: Erm... I was just joking in my first comment, but I disagree with your comment, because anyone who claims they use ZFC as their foundational system necessarily has made some very weird philosophical assumptions whether or not they know it.
$endgroup$
– user21820
Mar 10 at 14:10
1
$begingroup$
@YCor: But that's only if you think "truth within set theory" is meaningful. To refrain from prolonging this thread with our off-topic discussion, do you want to come to the logic chat-room?
$endgroup$
– user21820
Mar 10 at 14:16