If $Hleq G$ and $G-H$ is finite, then $G=H$ or $G$ is finite.
I'm trying to prove that if $Hleq G$ and $G-H$ is finite, then $G=H$ or $G$ is finite.
It is clear that if $|G-H|=0$ then $G=H$; otherwise, if $0<|G-H|=n<infty$ then $|G:H|leq|G-H|$.
The problem is that I can't find out how to use this fact to work out the proof.
Maybe this is not the right approach, so I would appreciate some hints, please.
group-theory finite-groups
edited Nov 29 '18 at 21:46
Shaun
8,820113681
asked Dec 2 '15 at 22:17
Jose PaterninaJose Paternina
660414
|
show 1 more comment
I'm trying to prove that if $Hleq G$ and $G-H$ is finite, then $G=H$ or $G$ is finite.
It is clear that if $|G-H|=0$ then $G=H$; otherwise, if $0<|G-H|=n<infty$ then $|G:H|leq|G-H|$.
The problem is that I can't find out how to use this fact to work out the proof.
Maybe this is not the right approach, so I would appreciate some hints, please.
group-theory finite-groups
edited Nov 29 '18 at 21:46
Shaun
8,820113681
asked Dec 2 '15 at 22:17
Jose PaterninaJose Paternina
660414
3
Use that $G$ is a union of cosets of $H$, and that $H$ is itself such a coset. Draw a picture.
– Myself
Dec 2 '15 at 22:19
1
@Myself Oh, ok. It wasn't that hard. Thanks!
– Jose Paternina
Dec 2 '15 at 22:27
Here is a discussion of what I have so far.
– Shaun
Nov 29 '18 at 20:52
1
@Shaun The complement contains representatives for all but one coset as well as at least one full coset. So there are finitely many cosets and they are all finite.
– Tobias Kildetoft
Nov 29 '18 at 21:39
Ah, I see! Thank you, @TobiasKildetoft. Would you mind posting that as an answer, please? It's just that it checks the question off the list of unanswered questions. If you don't want to, I'll copy & paste your comment as a community wiki answer (unless you have any objections) $ddotsmile$
– Shaun
Nov 29 '18 at 21:45
|
show 1 more comment
I'm trying to prove that if $Hleq G$ and $G-H$ is finite, then $G=H$ or $G$ is finite.
It is clear that if $|G-H|=0$ then $G=H$; otherwise, if $0<|G-H|=n<infty$ then $|G:H|leq|G-H|$.
The problem is that I can't find out how to use this fact to work out the proof.
Maybe this is not the right approach, so I would appreciate some hints, please.
group-theory finite-groups
edited Nov 29 '18 at 21:46
Shaun
8,820113681
asked Dec 2 '15 at 22:17
Jose PaterninaJose Paternina
660414
I'm trying to prove that if $Hleq G$ and $G-H$ is finite, then $G=H$ or $G$ is finite.
It is clear that if $|G-H|=0$ then $G=H$; otherwise, if $0<|G-H|=n<infty$ then $|G:H|leq|G-H|$.
The problem is that I can't find out how to use this fact to work out the proof.
Maybe this is not the right approach, so I would appreciate some hints, please.
group-theory finite-groups
group-theory finite-groups
edited Nov 29 '18 at 21:46
Shaun
8,820113681
asked Dec 2 '15 at 22:17
Jose PaterninaJose Paternina
660414
edited Nov 29 '18 at 21:46
Shaun
8,820113681
asked Dec 2 '15 at 22:17
Jose PaterninaJose Paternina
660414
edited Nov 29 '18 at 21:46
Shaun
8,820113681
edited Nov 29 '18 at 21:46
Shaun
8,820113681
edited Nov 29 '18 at 21:46
Shaun
8,820113681
8,820113681
asked Dec 2 '15 at 22:17
Jose PaterninaJose Paternina
660414
asked Dec 2 '15 at 22:17
Jose PaterninaJose Paternina
660414
asked Dec 2 '15 at 22:17
Jose PaterninaJose Paternina
660414
660414
3
Use that $G$ is a union of cosets of $H$, and that $H$ is itself such a coset. Draw a picture.
– Myself
Dec 2 '15 at 22:19
1
@Myself Oh, ok. It wasn't that hard. Thanks!
– Jose Paternina
Dec 2 '15 at 22:27
Here is a discussion of what I have so far.
– Shaun
Nov 29 '18 at 20:52
1
@Shaun The complement contains representatives for all but one coset as well as at least one full coset. So there are finitely many cosets and they are all finite.
– Tobias Kildetoft
Nov 29 '18 at 21:39
Ah, I see! Thank you, @TobiasKildetoft. Would you mind posting that as an answer, please? It's just that it checks the question off the list of unanswered questions. If you don't want to, I'll copy & paste your comment as a community wiki answer (unless you have any objections) $ddotsmile$
– Shaun
Nov 29 '18 at 21:45
|
show 1 more comment
3
Use that $G$ is a union of cosets of $H$, and that $H$ is itself such a coset. Draw a picture.
– Myself
Dec 2 '15 at 22:19
1
@Myself Oh, ok. It wasn't that hard. Thanks!
– Jose Paternina
Dec 2 '15 at 22:27
Here is a discussion of what I have so far.
– Shaun
Nov 29 '18 at 20:52
1
@Shaun The complement contains representatives for all but one coset as well as at least one full coset. So there are finitely many cosets and they are all finite.
– Tobias Kildetoft
Nov 29 '18 at 21:39
Ah, I see! Thank you, @TobiasKildetoft. Would you mind posting that as an answer, please? It's just that it checks the question off the list of unanswered questions. If you don't want to, I'll copy & paste your comment as a community wiki answer (unless you have any objections) $ddotsmile$
– Shaun
Nov 29 '18 at 21:45
3
3
Use that $G$ is a union of cosets of $H$, and that $H$ is itself such a coset. Draw a picture.
– Myself
Dec 2 '15 at 22:19
Use that $G$ is a union of cosets of $H$, and that $H$ is itself such a coset. Draw a picture.
– Myself
Dec 2 '15 at 22:19
1
1
@Myself Oh, ok. It wasn't that hard. Thanks!
– Jose Paternina
Dec 2 '15 at 22:27
@Myself Oh, ok. It wasn't that hard. Thanks!
– Jose Paternina
Dec 2 '15 at 22:27
Here is a discussion of what I have so far.
– Shaun
Nov 29 '18 at 20:52
Here is a discussion of what I have so far.
– Shaun
Nov 29 '18 at 20:52
1
1
@Shaun The complement contains representatives for all but one coset as well as at least one full coset. So there are finitely many cosets and they are all finite.
– Tobias Kildetoft
Nov 29 '18 at 21:39
@Shaun The complement contains representatives for all but one coset as well as at least one full coset. So there are finitely many cosets and they are all finite.
– Tobias Kildetoft
Nov 29 '18 at 21:39
Ah, I see! Thank you, @TobiasKildetoft. Would you mind posting that as an answer, please? It's just that it checks the question off the list of unanswered questions. If you don't want to, I'll copy & paste your comment as a community wiki answer (unless you have any objections) $ddotsmile$
– Shaun
Nov 29 '18 at 21:45
Ah, I see! Thank you, @TobiasKildetoft. Would you mind posting that as an answer, please? It's just that it checks the question off the list of unanswered questions. If you don't want to, I'll copy & paste your comment as a community wiki answer (unless you have any objections) $ddotsmile$
– Shaun
Nov 29 '18 at 21:45
|
show 1 more comment
1 Answer
1
active
oldest
votes
Let us assume that $Hneq G$ and show that then $G$ is finite.
Since the complement of $H$ in $G$ contains representatives of all but one of the cosets of $H$ in $G$ (missing only the coset $H$), there are finitely many cosets of $H$ in $G$.
But the complement also contains at least one full coset, so this coset is finite and hence all cosets are finite.
Since $G$ is the union of these finitely many finite cosets, it is itself finite.
answered Nov 30 '18 at 7:39
Tobias KildetoftTobias Kildetoft
16.7k14273
add a comment |
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Let us assume that $Hneq G$ and show that then $G$ is finite.
Since the complement of $H$ in $G$ contains representatives of all but one of the cosets of $H$ in $G$ (missing only the coset $H$), there are finitely many cosets of $H$ in $G$.
But the complement also contains at least one full coset, so this coset is finite and hence all cosets are finite.
Since $G$ is the union of these finitely many finite cosets, it is itself finite.
answered Nov 30 '18 at 7:39
Tobias KildetoftTobias Kildetoft
16.7k14273
add a comment |
Let us assume that $Hneq G$ and show that then $G$ is finite.
Since the complement of $H$ in $G$ contains representatives of all but one of the cosets of $H$ in $G$ (missing only the coset $H$), there are finitely many cosets of $H$ in $G$.
But the complement also contains at least one full coset, so this coset is finite and hence all cosets are finite.
Since $G$ is the union of these finitely many finite cosets, it is itself finite.
answered Nov 30 '18 at 7:39
Tobias KildetoftTobias Kildetoft
16.7k14273
add a comment |
Let us assume that $Hneq G$ and show that then $G$ is finite.
Since the complement of $H$ in $G$ contains representatives of all but one of the cosets of $H$ in $G$ (missing only the coset $H$), there are finitely many cosets of $H$ in $G$.
But the complement also contains at least one full coset, so this coset is finite and hence all cosets are finite.
Since $G$ is the union of these finitely many finite cosets, it is itself finite.
answered Nov 30 '18 at 7:39
Tobias KildetoftTobias Kildetoft
16.7k14273
Let us assume that $Hneq G$ and show that then $G$ is finite.
Since the complement of $H$ in $G$ contains representatives of all but one of the cosets of $H$ in $G$ (missing only the coset $H$), there are finitely many cosets of $H$ in $G$.
But the complement also contains at least one full coset, so this coset is finite and hence all cosets are finite.
Since $G$ is the union of these finitely many finite cosets, it is itself finite.
answered Nov 30 '18 at 7:39
Tobias KildetoftTobias Kildetoft
16.7k14273
answered Nov 30 '18 at 7:39
Tobias KildetoftTobias Kildetoft
16.7k14273
answered Nov 30 '18 at 7:39
Tobias KildetoftTobias Kildetoft
16.7k14273
answered Nov 30 '18 at 7:39
Tobias KildetoftTobias Kildetoft
16.7k14273
16.7k14273
add a comment |
add a comment |
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3
Use that $G$ is a union of cosets of $H$, and that $H$ is itself such a coset. Draw a picture.
– Myself
Dec 2 '15 at 22:19
1
@Myself Oh, ok. It wasn't that hard. Thanks!
– Jose Paternina
Dec 2 '15 at 22:27
Here is a discussion of what I have so far.
– Shaun
Nov 29 '18 at 20:52
1
@Shaun The complement contains representatives for all but one coset as well as at least one full coset. So there are finitely many cosets and they are all finite.
– Tobias Kildetoft
Nov 29 '18 at 21:39
Ah, I see! Thank you, @TobiasKildetoft. Would you mind posting that as an answer, please? It's just that it checks the question off the list of unanswered questions. If you don't want to, I'll copy & paste your comment as a community wiki answer (unless you have any objections) $ddotsmile$
– Shaun
Nov 29 '18 at 21:45