Discrete subgroups of $O(2)$ are all finite. [closed]
$begingroup$
How to prove that discrete subgroups of the orthogonal group in dimension $2$ are all of finite order?
Please help me in this regard. Thank you very much.
group-theory
edited Jan 1 at 5:58
Kemono Chen
2,9701739
asked Jan 1 at 5:46
Dbchatto67Dbchatto67
536116
$endgroup$
closed as off-topic by Mike Miller, José Carlos Santos, Shaun, Saad, ancientmathematician Jan 1 at 16:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Mike Miller, José Carlos Santos, Shaun, Saad, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How to prove that discrete subgroups of the orthogonal group in dimension $2$ are all of finite order?
Please help me in this regard. Thank you very much.
group-theory
edited Jan 1 at 5:58
Kemono Chen
2,9701739
asked Jan 1 at 5:46
Dbchatto67Dbchatto67
536116
$endgroup$
closed as off-topic by Mike Miller, José Carlos Santos, Shaun, Saad, ancientmathematician Jan 1 at 16:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Mike Miller, José Carlos Santos, Shaun, Saad, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How to prove that discrete subgroups of the orthogonal group in dimension $2$ are all of finite order?
Please help me in this regard. Thank you very much.
group-theory
edited Jan 1 at 5:58
Kemono Chen
2,9701739
asked Jan 1 at 5:46
Dbchatto67Dbchatto67
536116
$endgroup$
How to prove that discrete subgroups of the orthogonal group in dimension $2$ are all of finite order?
Please help me in this regard. Thank you very much.
group-theory
group-theory
edited Jan 1 at 5:58
Kemono Chen
2,9701739
asked Jan 1 at 5:46
Dbchatto67Dbchatto67
536116
edited Jan 1 at 5:58
Kemono Chen
2,9701739
asked Jan 1 at 5:46
Dbchatto67Dbchatto67
536116
edited Jan 1 at 5:58
Kemono Chen
2,9701739
edited Jan 1 at 5:58
Kemono Chen
2,9701739
edited Jan 1 at 5:58
Kemono Chen
2,9701739
2,9701739
asked Jan 1 at 5:46
Dbchatto67Dbchatto67
536116
asked Jan 1 at 5:46
Dbchatto67Dbchatto67
536116
asked Jan 1 at 5:46
Dbchatto67Dbchatto67
536116
536116
closed as off-topic by Mike Miller, José Carlos Santos, Shaun, Saad, ancientmathematician Jan 1 at 16:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Mike Miller, José Carlos Santos, Shaun, Saad, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Mike Miller, José Carlos Santos, Shaun, Saad, ancientmathematician Jan 1 at 16:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Mike Miller, José Carlos Santos, Shaun, Saad, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It suffices to note that $O(2)$ is compact.
Suppose that $G$ is an infinite subgroup of $O(2)$. Then $G$ contains an infinite sequence $(g_k)_{k in Bbb N}$. Because $O(2)$ is compact, this sequence contains a convergent subsequence, so suppose WLOG that $g_k to g in O(2)$.
Verify that the induced topology on ${g_k}_{k in Bbb N}$ is not the discrete topology. In particular, note that any open subset of $O(2)$ containing $g$ must contain all but finitely many elements of the sequence.
answered Jan 1 at 6:00
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
$begingroup$
You use Bolzano-Weierstrass theorem. Right?
$endgroup$
– Dbchatto67
Jan 1 at 6:03
1
$begingroup$
But how do I prove that $O(2)$ is compact?
$endgroup$
– Dbchatto67
Jan 1 at 6:14
3
$begingroup$
By Heine-Borel, it suffices to note that $O(2)$ is a closed and bounded subset of $Bbb R^{2 times 2}$
$endgroup$
– Omnomnomnom
Jan 1 at 6:18
1
$begingroup$
So by using Heine-Borel theorem it follows $O(2)$ is compact.
$endgroup$
– Dbchatto67
Jan 1 at 7:32
1
$begingroup$
@Dbchatto I am glad to see you complete the proof in the comments! It's always good to see that.
$endgroup$
– Mike Miller
Jan 1 at 8:44
|
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It suffices to note that $O(2)$ is compact.
Suppose that $G$ is an infinite subgroup of $O(2)$. Then $G$ contains an infinite sequence $(g_k)_{k in Bbb N}$. Because $O(2)$ is compact, this sequence contains a convergent subsequence, so suppose WLOG that $g_k to g in O(2)$.
Verify that the induced topology on ${g_k}_{k in Bbb N}$ is not the discrete topology. In particular, note that any open subset of $O(2)$ containing $g$ must contain all but finitely many elements of the sequence.
answered Jan 1 at 6:00
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
$begingroup$
You use Bolzano-Weierstrass theorem. Right?
$endgroup$
– Dbchatto67
Jan 1 at 6:03
1
$begingroup$
But how do I prove that $O(2)$ is compact?
$endgroup$
– Dbchatto67
Jan 1 at 6:14
3
$begingroup$
By Heine-Borel, it suffices to note that $O(2)$ is a closed and bounded subset of $Bbb R^{2 times 2}$
$endgroup$
– Omnomnomnom
Jan 1 at 6:18
1
$begingroup$
So by using Heine-Borel theorem it follows $O(2)$ is compact.
$endgroup$
– Dbchatto67
Jan 1 at 7:32
1
$begingroup$
@Dbchatto I am glad to see you complete the proof in the comments! It's always good to see that.
$endgroup$
– Mike Miller
Jan 1 at 8:44
|
show 5 more comments
$begingroup$
It suffices to note that $O(2)$ is compact.
Suppose that $G$ is an infinite subgroup of $O(2)$. Then $G$ contains an infinite sequence $(g_k)_{k in Bbb N}$. Because $O(2)$ is compact, this sequence contains a convergent subsequence, so suppose WLOG that $g_k to g in O(2)$.
Verify that the induced topology on ${g_k}_{k in Bbb N}$ is not the discrete topology. In particular, note that any open subset of $O(2)$ containing $g$ must contain all but finitely many elements of the sequence.
answered Jan 1 at 6:00
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
$begingroup$
You use Bolzano-Weierstrass theorem. Right?
$endgroup$
– Dbchatto67
Jan 1 at 6:03
1
$begingroup$
But how do I prove that $O(2)$ is compact?
$endgroup$
– Dbchatto67
Jan 1 at 6:14
3
$begingroup$
By Heine-Borel, it suffices to note that $O(2)$ is a closed and bounded subset of $Bbb R^{2 times 2}$
$endgroup$
– Omnomnomnom
Jan 1 at 6:18
1
$begingroup$
So by using Heine-Borel theorem it follows $O(2)$ is compact.
$endgroup$
– Dbchatto67
Jan 1 at 7:32
1
$begingroup$
@Dbchatto I am glad to see you complete the proof in the comments! It's always good to see that.
$endgroup$
– Mike Miller
Jan 1 at 8:44
|
show 5 more comments
$begingroup$
It suffices to note that $O(2)$ is compact.
Suppose that $G$ is an infinite subgroup of $O(2)$. Then $G$ contains an infinite sequence $(g_k)_{k in Bbb N}$. Because $O(2)$ is compact, this sequence contains a convergent subsequence, so suppose WLOG that $g_k to g in O(2)$.
Verify that the induced topology on ${g_k}_{k in Bbb N}$ is not the discrete topology. In particular, note that any open subset of $O(2)$ containing $g$ must contain all but finitely many elements of the sequence.
answered Jan 1 at 6:00
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
It suffices to note that $O(2)$ is compact.
Suppose that $G$ is an infinite subgroup of $O(2)$. Then $G$ contains an infinite sequence $(g_k)_{k in Bbb N}$. Because $O(2)$ is compact, this sequence contains a convergent subsequence, so suppose WLOG that $g_k to g in O(2)$.
Verify that the induced topology on ${g_k}_{k in Bbb N}$ is not the discrete topology. In particular, note that any open subset of $O(2)$ containing $g$ must contain all but finitely many elements of the sequence.
answered Jan 1 at 6:00
OmnomnomnomOmnomnomnom
127k790178
edited Jan 1 at 6:11
answered Jan 1 at 6:00
OmnomnomnomOmnomnomnom
127k790178
answered Jan 1 at 6:00
OmnomnomnomOmnomnomnom
127k790178
answered Jan 1 at 6:00
OmnomnomnomOmnomnomnom
127k790178
127k790178
$begingroup$
You use Bolzano-Weierstrass theorem. Right?
$endgroup$
– Dbchatto67
Jan 1 at 6:03
1
$begingroup$
But how do I prove that $O(2)$ is compact?
$endgroup$
– Dbchatto67
Jan 1 at 6:14
3
$begingroup$
By Heine-Borel, it suffices to note that $O(2)$ is a closed and bounded subset of $Bbb R^{2 times 2}$
$endgroup$
– Omnomnomnom
Jan 1 at 6:18
1
$begingroup$
So by using Heine-Borel theorem it follows $O(2)$ is compact.
$endgroup$
– Dbchatto67
Jan 1 at 7:32
1
$begingroup$
@Dbchatto I am glad to see you complete the proof in the comments! It's always good to see that.
$endgroup$
– Mike Miller
Jan 1 at 8:44
|
show 5 more comments
$begingroup$
You use Bolzano-Weierstrass theorem. Right?
$endgroup$
– Dbchatto67
Jan 1 at 6:03
1
$begingroup$
But how do I prove that $O(2)$ is compact?
$endgroup$
– Dbchatto67
Jan 1 at 6:14
3
$begingroup$
By Heine-Borel, it suffices to note that $O(2)$ is a closed and bounded subset of $Bbb R^{2 times 2}$
$endgroup$
– Omnomnomnom
Jan 1 at 6:18
1
$begingroup$
So by using Heine-Borel theorem it follows $O(2)$ is compact.
$endgroup$
– Dbchatto67
Jan 1 at 7:32
1
$begingroup$
@Dbchatto I am glad to see you complete the proof in the comments! It's always good to see that.
$endgroup$
– Mike Miller
Jan 1 at 8:44
$begingroup$
You use Bolzano-Weierstrass theorem. Right?
$endgroup$
– Dbchatto67
Jan 1 at 6:03
$begingroup$
You use Bolzano-Weierstrass theorem. Right?
$endgroup$
– Dbchatto67
Jan 1 at 6:03
1
1
$begingroup$
But how do I prove that $O(2)$ is compact?
$endgroup$
– Dbchatto67
Jan 1 at 6:14
$begingroup$
But how do I prove that $O(2)$ is compact?
$endgroup$
– Dbchatto67
Jan 1 at 6:14
3
3
$begingroup$
By Heine-Borel, it suffices to note that $O(2)$ is a closed and bounded subset of $Bbb R^{2 times 2}$
$endgroup$
– Omnomnomnom
Jan 1 at 6:18
$begingroup$
By Heine-Borel, it suffices to note that $O(2)$ is a closed and bounded subset of $Bbb R^{2 times 2}$
$endgroup$
– Omnomnomnom
Jan 1 at 6:18
1
1
$begingroup$
So by using Heine-Borel theorem it follows $O(2)$ is compact.
$endgroup$
– Dbchatto67
Jan 1 at 7:32
$begingroup$
So by using Heine-Borel theorem it follows $O(2)$ is compact.
$endgroup$
– Dbchatto67
Jan 1 at 7:32
1
1
$begingroup$
@Dbchatto I am glad to see you complete the proof in the comments! It's always good to see that.
$endgroup$
– Mike Miller
Jan 1 at 8:44
$begingroup$
@Dbchatto I am glad to see you complete the proof in the comments! It's always good to see that.
$endgroup$
– Mike Miller
Jan 1 at 8:44
|
show 5 more comments
