Group containing both $5mathbbZ$ and $7mathbbZ$

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

Prove that there doesnot exist any proper subgroup of $(mathbbZ,+)$ containing both $5mathbbZ$ and $7mathbbZ$
Since $5$ and $7$ are coprime hence there cannot exist any proper subgroup containing multiples of both $5$ and $7$ if it contains then it must be the group $mathbbZ$ itself
Is m above reason correct please help?

abstract-algebra

share|cite|improve this question

edited Jul 15 at 12:33

cansomeonehelpmeout

4,7593830

asked Jul 15 at 12:25

Shrimon Mukherjee

195

add a comment | 

up vote
2
down vote

favorite

Prove that there doesnot exist any proper subgroup of $(mathbbZ,+)$ containing both $5mathbbZ$ and $7mathbbZ$
Since $5$ and $7$ are coprime hence there cannot exist any proper subgroup containing multiples of both $5$ and $7$ if it contains then it must be the group $mathbbZ$ itself
Is m above reason correct please help?

abstract-algebra

share|cite|improve this question

edited Jul 15 at 12:33

cansomeonehelpmeout

4,7593830

asked Jul 15 at 12:25

Shrimon Mukherjee

195

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Prove that there doesnot exist any proper subgroup of $(mathbbZ,+)$ containing both $5mathbbZ$ and $7mathbbZ$
Since $5$ and $7$ are coprime hence there cannot exist any proper subgroup containing multiples of both $5$ and $7$ if it contains then it must be the group $mathbbZ$ itself
Is m above reason correct please help?

abstract-algebra

share|cite|improve this question

edited Jul 15 at 12:33

cansomeonehelpmeout

4,7593830

asked Jul 15 at 12:25

Shrimon Mukherjee

195

Prove that there doesnot exist any proper subgroup of $(mathbbZ,+)$ containing both $5mathbbZ$ and $7mathbbZ$
Since $5$ and $7$ are coprime hence there cannot exist any proper subgroup containing multiples of both $5$ and $7$ if it contains then it must be the group $mathbbZ$ itself
Is m above reason correct please help?

abstract-algebra

share|cite|improve this question

edited Jul 15 at 12:33

cansomeonehelpmeout

4,7593830

asked Jul 15 at 12:25

Shrimon Mukherjee

195

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 12:33

cansomeonehelpmeout

4,7593830

edited Jul 15 at 12:33

cansomeonehelpmeout

4,7593830

edited Jul 15 at 12:33

cansomeonehelpmeout

4,7593830

4,7593830

asked Jul 15 at 12:25

Shrimon Mukherjee

195

asked Jul 15 at 12:25

Shrimon Mukherjee

195

asked Jul 15 at 12:25

Shrimon Mukherjee

195

195

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
5
down vote

You've basically just rephrased the problem, throwing in the word coprime, which is essential, but you haven't really shown how. Unless you have a theorem somewhere saying something like "No proper subgroup of $(Bbb Z,+)$ can contain two coprime elements", I wouldn't find that enough.

Try this: show specifically why a subgroup containing both $5$ and $7$ must contain $1$.

share|cite|improve this answer

answered Jul 15 at 12:28

Arthur

99k793175

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why should it contains 1
  – Shrimon Mukherjee
  Jul 15 at 12:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ShrimonMukherjee List all the elements between $0$ and $20$ which you can prove that the group contains, using that the group contains $5$ and $7$. You ought to be able to get to $1$ that way.
  – Arthur
  Jul 15 at 12:40
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

If a subgroup $G le (mathbbZ, +)$ contains $5mathbbZ$ and $7mathbbZ$ then it also contains

$$1 = 50 - 49 = 10cdot 5 - 7 cdot 7$$

so $G = mathbbZ$.

share|cite|improve this answer

answered Jul 15 at 13:57

mechanodroid

22.3k52041

add a comment | 

up vote
0
down vote

The key is that 5 and 7 are indeed coprime, but the numbers themselves are irrelevant. In fact, you might see it clearer if you turn to a more general situation:

Claim. If $m,n in mathbbZ$ are two coprime integers, then the only subgroup of $(mathbbZ,+)$ containing $mmathbbZ$ and $nmathbbZ$ is precisely $mathbbZ$.

Proof. Let $H subseteq mathbbZ$ be a subgrup of $(mathbbZ,+)$ containing $mmathbbZ$ and $nmathbbZ$. Clearly $mathbbZ=1mathbbZ$, so it suffices to show that $1 in mathbbZ$.

Since $m$ and $n$ are coprime, by Bézout's identity we have that
$$mx+ny=1$$
for some $x,y in mathbbZ$.
Besides, $m,n in H$ and $H$ is closed under addition, so it follows that $1 in H$ and we are done.

share|cite|improve this answer

answered Jul 16 at 19:41

Don

1263

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852482%2fgroup-containing-both-5-mathbbz-and-7-mathbbz%23new-answer', 'question_page');

);

Post as a guest

Name Email

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote

You've basically just rephrased the problem, throwing in the word coprime, which is essential, but you haven't really shown how. Unless you have a theorem somewhere saying something like "No proper subgroup of $(Bbb Z,+)$ can contain two coprime elements", I wouldn't find that enough.

Try this: show specifically why a subgroup containing both $5$ and $7$ must contain $1$.

share|cite|improve this answer

answered Jul 15 at 12:28

Arthur

99k793175

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why should it contains 1
  – Shrimon Mukherjee
  Jul 15 at 12:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ShrimonMukherjee List all the elements between $0$ and $20$ which you can prove that the group contains, using that the group contains $5$ and $7$. You ought to be able to get to $1$ that way.
  – Arthur
  Jul 15 at 12:40
  
  

  
  
  
  

add a comment | 

up vote
5
down vote

You've basically just rephrased the problem, throwing in the word coprime, which is essential, but you haven't really shown how. Unless you have a theorem somewhere saying something like "No proper subgroup of $(Bbb Z,+)$ can contain two coprime elements", I wouldn't find that enough.

Try this: show specifically why a subgroup containing both $5$ and $7$ must contain $1$.

share|cite|improve this answer

answered Jul 15 at 12:28

Arthur

99k793175

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why should it contains 1
  – Shrimon Mukherjee
  Jul 15 at 12:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ShrimonMukherjee List all the elements between $0$ and $20$ which you can prove that the group contains, using that the group contains $5$ and $7$. You ought to be able to get to $1$ that way.
  – Arthur
  Jul 15 at 12:40
  
  

  
  
  
  

add a comment | 

up vote
5
down vote

up vote
5
down vote

You've basically just rephrased the problem, throwing in the word coprime, which is essential, but you haven't really shown how. Unless you have a theorem somewhere saying something like "No proper subgroup of $(Bbb Z,+)$ can contain two coprime elements", I wouldn't find that enough.

Try this: show specifically why a subgroup containing both $5$ and $7$ must contain $1$.

share|cite|improve this answer

answered Jul 15 at 12:28

Arthur

99k793175

You've basically just rephrased the problem, throwing in the word coprime, which is essential, but you haven't really shown how. Unless you have a theorem somewhere saying something like "No proper subgroup of $(Bbb Z,+)$ can contain two coprime elements", I wouldn't find that enough.

Try this: show specifically why a subgroup containing both $5$ and $7$ must contain $1$.

share|cite|improve this answer

answered Jul 15 at 12:28

Arthur

99k793175

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 12:28

Arthur

99k793175

answered Jul 15 at 12:28

Arthur

99k793175

answered Jul 15 at 12:28

Arthur

99k793175

99k793175

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why should it contains 1
  – Shrimon Mukherjee
  Jul 15 at 12:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ShrimonMukherjee List all the elements between $0$ and $20$ which you can prove that the group contains, using that the group contains $5$ and $7$. You ought to be able to get to $1$ that way.
  – Arthur
  Jul 15 at 12:40
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why should it contains 1
  – Shrimon Mukherjee
  Jul 15 at 12:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ShrimonMukherjee List all the elements between $0$ and $20$ which you can prove that the group contains, using that the group contains $5$ and $7$. You ought to be able to get to $1$ that way.
  – Arthur
  Jul 15 at 12:40
  
  

  
  
  
  

Why should it contains 1
– Shrimon Mukherjee
Jul 15 at 12:37

Why should it contains 1
– Shrimon Mukherjee
Jul 15 at 12:37

1

1

@ShrimonMukherjee List all the elements between $0$ and $20$ which you can prove that the group contains, using that the group contains $5$ and $7$. You ought to be able to get to $1$ that way.
– Arthur
Jul 15 at 12:40

@ShrimonMukherjee List all the elements between $0$ and $20$ which you can prove that the group contains, using that the group contains $5$ and $7$. You ought to be able to get to $1$ that way.
– Arthur
Jul 15 at 12:40

add a comment | 

up vote
3
down vote

If a subgroup $G le (mathbbZ, +)$ contains $5mathbbZ$ and $7mathbbZ$ then it also contains

$$1 = 50 - 49 = 10cdot 5 - 7 cdot 7$$

so $G = mathbbZ$.

share|cite|improve this answer

answered Jul 15 at 13:57

mechanodroid

22.3k52041

add a comment | 

up vote
3
down vote

If a subgroup $G le (mathbbZ, +)$ contains $5mathbbZ$ and $7mathbbZ$ then it also contains

$$1 = 50 - 49 = 10cdot 5 - 7 cdot 7$$

so $G = mathbbZ$.

share|cite|improve this answer

answered Jul 15 at 13:57

mechanodroid

22.3k52041

add a comment | 

up vote
3
down vote

up vote
3
down vote

If a subgroup $G le (mathbbZ, +)$ contains $5mathbbZ$ and $7mathbbZ$ then it also contains

$$1 = 50 - 49 = 10cdot 5 - 7 cdot 7$$

so $G = mathbbZ$.

share|cite|improve this answer

answered Jul 15 at 13:57

mechanodroid

22.3k52041

If a subgroup $G le (mathbbZ, +)$ contains $5mathbbZ$ and $7mathbbZ$ then it also contains

$$1 = 50 - 49 = 10cdot 5 - 7 cdot 7$$

so $G = mathbbZ$.

share|cite|improve this answer

answered Jul 15 at 13:57

mechanodroid

22.3k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 13:57

mechanodroid

22.3k52041

answered Jul 15 at 13:57

mechanodroid

22.3k52041

answered Jul 15 at 13:57

mechanodroid

22.3k52041

22.3k52041

add a comment | 

add a comment | 

up vote
0
down vote

The key is that 5 and 7 are indeed coprime, but the numbers themselves are irrelevant. In fact, you might see it clearer if you turn to a more general situation:

Claim. If $m,n in mathbbZ$ are two coprime integers, then the only subgroup of $(mathbbZ,+)$ containing $mmathbbZ$ and $nmathbbZ$ is precisely $mathbbZ$.

Proof. Let $H subseteq mathbbZ$ be a subgrup of $(mathbbZ,+)$ containing $mmathbbZ$ and $nmathbbZ$. Clearly $mathbbZ=1mathbbZ$, so it suffices to show that $1 in mathbbZ$.

Since $m$ and $n$ are coprime, by Bézout's identity we have that
$$mx+ny=1$$
for some $x,y in mathbbZ$.
Besides, $m,n in H$ and $H$ is closed under addition, so it follows that $1 in H$ and we are done.

share|cite|improve this answer

answered Jul 16 at 19:41

Don

1263

add a comment | 

up vote
0
down vote

The key is that 5 and 7 are indeed coprime, but the numbers themselves are irrelevant. In fact, you might see it clearer if you turn to a more general situation:

Claim. If $m,n in mathbbZ$ are two coprime integers, then the only subgroup of $(mathbbZ,+)$ containing $mmathbbZ$ and $nmathbbZ$ is precisely $mathbbZ$.

Proof. Let $H subseteq mathbbZ$ be a subgrup of $(mathbbZ,+)$ containing $mmathbbZ$ and $nmathbbZ$. Clearly $mathbbZ=1mathbbZ$, so it suffices to show that $1 in mathbbZ$.

Since $m$ and $n$ are coprime, by Bézout's identity we have that
$$mx+ny=1$$
for some $x,y in mathbbZ$.
Besides, $m,n in H$ and $H$ is closed under addition, so it follows that $1 in H$ and we are done.

share|cite|improve this answer

answered Jul 16 at 19:41

Don

1263

add a comment | 

up vote
0
down vote

up vote
0
down vote

The key is that 5 and 7 are indeed coprime, but the numbers themselves are irrelevant. In fact, you might see it clearer if you turn to a more general situation:

Claim. If $m,n in mathbbZ$ are two coprime integers, then the only subgroup of $(mathbbZ,+)$ containing $mmathbbZ$ and $nmathbbZ$ is precisely $mathbbZ$.

Proof. Let $H subseteq mathbbZ$ be a subgrup of $(mathbbZ,+)$ containing $mmathbbZ$ and $nmathbbZ$. Clearly $mathbbZ=1mathbbZ$, so it suffices to show that $1 in mathbbZ$.

Since $m$ and $n$ are coprime, by Bézout's identity we have that
$$mx+ny=1$$
for some $x,y in mathbbZ$.
Besides, $m,n in H$ and $H$ is closed under addition, so it follows that $1 in H$ and we are done.

share|cite|improve this answer

answered Jul 16 at 19:41

Don

1263

The key is that 5 and 7 are indeed coprime, but the numbers themselves are irrelevant. In fact, you might see it clearer if you turn to a more general situation:

Claim. If $m,n in mathbbZ$ are two coprime integers, then the only subgroup of $(mathbbZ,+)$ containing $mmathbbZ$ and $nmathbbZ$ is precisely $mathbbZ$.

Proof. Let $H subseteq mathbbZ$ be a subgrup of $(mathbbZ,+)$ containing $mmathbbZ$ and $nmathbbZ$. Clearly $mathbbZ=1mathbbZ$, so it suffices to show that $1 in mathbbZ$.

Since $m$ and $n$ are coprime, by Bézout's identity we have that
$$mx+ny=1$$
for some $x,y in mathbbZ$.
Besides, $m,n in H$ and $H$ is closed under addition, so it follows that $1 in H$ and we are done.

share|cite|improve this answer

answered Jul 16 at 19:41

Don

1263

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 19:41

Don

1263

answered Jul 16 at 19:41

Don

1263

answered Jul 16 at 19:41

Don

1263

1263

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852482%2fgroup-containing-both-5-mathbbz-and-7-mathbbz%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email