Is subgroup of index 2 in any group is unique?

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

I know that subgroup of index 2 is normal .

I am interested in knowing that is that subgroup is unqiue or there exist example of subgroup which can have 2 subgroup of index 2?

If there is example exist , then what is condition on subgroup implies that subgroup of index 2 is unique?

Any Help will be appreciated.

abstract-algebra group-theory normal-subgroups

share|cite|improve this question

asked 1 hour ago

user594188

785

New contributor

user594188 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
2
down vote

favorite

I know that subgroup of index 2 is normal .

I am interested in knowing that is that subgroup is unqiue or there exist example of subgroup which can have 2 subgroup of index 2?

If there is example exist , then what is condition on subgroup implies that subgroup of index 2 is unique?

Any Help will be appreciated.

abstract-algebra group-theory normal-subgroups

share|cite|improve this question

asked 1 hour ago

user594188

785

New contributor

user594188 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I know that subgroup of index 2 is normal .

I am interested in knowing that is that subgroup is unqiue or there exist example of subgroup which can have 2 subgroup of index 2?

If there is example exist , then what is condition on subgroup implies that subgroup of index 2 is unique?

Any Help will be appreciated.

abstract-algebra group-theory normal-subgroups

share|cite|improve this question

asked 1 hour ago

user594188

785

New contributor

user594188 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

I know that subgroup of index 2 is normal .

I am interested in knowing that is that subgroup is unqiue or there exist example of subgroup which can have 2 subgroup of index 2?

If there is example exist , then what is condition on subgroup implies that subgroup of index 2 is unique?

Any Help will be appreciated.

abstract-algebra group-theory normal-subgroups

abstract-algebra group-theory normal-subgroups

share|cite|improve this question

asked 1 hour ago

user594188

785

New contributor

user594188 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 1 hour ago

user594188

785

New contributor

user594188 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

share|cite|improve this question

asked 1 hour ago

user594188

785

New contributor

user594188 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

user594188

785

asked 1 hour ago

user594188

785

785

New contributor

user594188 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

user594188 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

user594188 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
5
down vote



accepted

Unfortunately it is not even unique up to isomorphism! Take, for instance, $(mathbbZ/2mathbbZ)times (mathbbZ/4mathbbZ)$.

share|cite|improve this answer

answered 1 hour ago

ThomasGrubb

9,95011335

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good example. Another: the dihedral group $D_4$ of order $8$ has at least two subgroups of order $4$: the Klein $4$-group and the subgroup consisting of the rotations. They are also non-iso.
  – Randall
  1 hour ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

No way is it unique (generally). The group $G=mathbbZ_2 times mathbbZ_2$ has three subgroups of index $2$: $mathbbZ_2 times0$, the vice versa, and $(0,0), (1,1)$. You can use the same idea to cook up plenty of other examples.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Randall

7,5321925

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
);



);

user594188 is a new contributor. Be nice, and check out our Code of Conduct.

 

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%2f2922390%2fis-subgroup-of-index-2-in-any-group-is-unique%23new-answer', 'question_page');

);

Post as a guest

Name Email

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

Unfortunately it is not even unique up to isomorphism! Take, for instance, $(mathbbZ/2mathbbZ)times (mathbbZ/4mathbbZ)$.

share|cite|improve this answer

answered 1 hour ago

ThomasGrubb

9,95011335

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good example. Another: the dihedral group $D_4$ of order $8$ has at least two subgroups of order $4$: the Klein $4$-group and the subgroup consisting of the rotations. They are also non-iso.
  – Randall
  1 hour ago
  
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

Unfortunately it is not even unique up to isomorphism! Take, for instance, $(mathbbZ/2mathbbZ)times (mathbbZ/4mathbbZ)$.

share|cite|improve this answer

answered 1 hour ago

ThomasGrubb

9,95011335

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good example. Another: the dihedral group $D_4$ of order $8$ has at least two subgroups of order $4$: the Klein $4$-group and the subgroup consisting of the rotations. They are also non-iso.
  – Randall
  1 hour ago
  
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

Unfortunately it is not even unique up to isomorphism! Take, for instance, $(mathbbZ/2mathbbZ)times (mathbbZ/4mathbbZ)$.

share|cite|improve this answer

answered 1 hour ago

ThomasGrubb

9,95011335

Unfortunately it is not even unique up to isomorphism! Take, for instance, $(mathbbZ/2mathbbZ)times (mathbbZ/4mathbbZ)$.

share|cite|improve this answer

answered 1 hour ago

ThomasGrubb

9,95011335

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

ThomasGrubb

9,95011335

answered 1 hour ago

ThomasGrubb

9,95011335

answered 1 hour ago

ThomasGrubb

9,95011335

9,95011335

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good example. Another: the dihedral group $D_4$ of order $8$ has at least two subgroups of order $4$: the Klein $4$-group and the subgroup consisting of the rotations. They are also non-iso.
  – Randall
  1 hour ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good example. Another: the dihedral group $D_4$ of order $8$ has at least two subgroups of order $4$: the Klein $4$-group and the subgroup consisting of the rotations. They are also non-iso.
  – Randall
  1 hour ago
  
  

  
  
  
  

Good example. Another: the dihedral group $D_4$ of order $8$ has at least two subgroups of order $4$: the Klein $4$-group and the subgroup consisting of the rotations. They are also non-iso.
– Randall
1 hour ago

Good example. Another: the dihedral group $D_4$ of order $8$ has at least two subgroups of order $4$: the Klein $4$-group and the subgroup consisting of the rotations. They are also non-iso.
– Randall
1 hour ago

add a comment | 

up vote
1
down vote

No way is it unique (generally). The group $G=mathbbZ_2 times mathbbZ_2$ has three subgroups of index $2$: $mathbbZ_2 times0$, the vice versa, and $(0,0), (1,1)$. You can use the same idea to cook up plenty of other examples.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Randall

7,5321925

add a comment | 

up vote
1
down vote

No way is it unique (generally). The group $G=mathbbZ_2 times mathbbZ_2$ has three subgroups of index $2$: $mathbbZ_2 times0$, the vice versa, and $(0,0), (1,1)$. You can use the same idea to cook up plenty of other examples.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Randall

7,5321925

add a comment | 

up vote
1
down vote

up vote
1
down vote

No way is it unique (generally). The group $G=mathbbZ_2 times mathbbZ_2$ has three subgroups of index $2$: $mathbbZ_2 times0$, the vice versa, and $(0,0), (1,1)$. You can use the same idea to cook up plenty of other examples.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Randall

7,5321925

No way is it unique (generally). The group $G=mathbbZ_2 times mathbbZ_2$ has three subgroups of index $2$: $mathbbZ_2 times0$, the vice versa, and $(0,0), (1,1)$. You can use the same idea to cook up plenty of other examples.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Randall

7,5321925

share|cite|improve this answer

share|cite|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 1 hour ago

Randall

7,5321925

answered 1 hour ago

Randall

7,5321925

answered 1 hour ago

Randall

7,5321925

7,5321925

add a comment | 

add a comment | 

user594188 is a new contributor. Be nice, and check out our Code of Conduct.

 

draft saved

draft discarded

user594188 is a new contributor. Be nice, and check out our Code of Conduct.

user594188 is a new contributor. Be nice, and check out our Code of Conduct.

user594188 is a new contributor. Be nice, and check out our Code of Conduct.

 

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%2f2922390%2fis-subgroup-of-index-2-in-any-group-is-unique%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