Example of two subgroups $H$ and $K$ of a non-abelian group $G$ such that $HK$ is not a subgroup of $G$? [on hold]

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I need to disprove that the set $HK$ is not a subgroup of $G$ if $G$ is non-abelian. Does anyone have a trivial or easy counterexample? I cannot think of one...

group-theory abelian-groups

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asked Aug 4 at 1:02

numericalorange

1,147110

put on hold as off-topic by Derek Holt, Jyrki Lahtonen, amWhy, José Carlos Santos, John Ma 2 days ago

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Jyrki Lahtonen, amWhy, José Carlos Santos, John Ma

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you men is there nonabelian $G$ and subgroups $H,K$ for which $HK$ is not a subgroup? [Your use of double negative made me ask]
  – coffeemath
  Aug 4 at 1:07
  
  

  
  
  
  

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up vote
0
down vote

favorite

2

I need to disprove that the set $HK$ is not a subgroup of $G$ if $G$ is non-abelian. Does anyone have a trivial or easy counterexample? I cannot think of one...

group-theory abelian-groups

share|cite|improve this question

asked Aug 4 at 1:02

numericalorange

1,147110

put on hold as off-topic by Derek Holt, Jyrki Lahtonen, amWhy, José Carlos Santos, John Ma 2 days ago

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Jyrki Lahtonen, amWhy, José Carlos Santos, John Ma

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you men is there nonabelian $G$ and subgroups $H,K$ for which $HK$ is not a subgroup? [Your use of double negative made me ask]
  – coffeemath
  Aug 4 at 1:07
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

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2

up vote
0
down vote

favorite

2

2

I need to disprove that the set $HK$ is not a subgroup of $G$ if $G$ is non-abelian. Does anyone have a trivial or easy counterexample? I cannot think of one...

group-theory abelian-groups

share|cite|improve this question

asked Aug 4 at 1:02

numericalorange

1,147110

I need to disprove that the set $HK$ is not a subgroup of $G$ if $G$ is non-abelian. Does anyone have a trivial or easy counterexample? I cannot think of one...

group-theory abelian-groups

share|cite|improve this question

asked Aug 4 at 1:02

numericalorange

1,147110

share|cite|improve this question

share|cite|improve this question

asked Aug 4 at 1:02

numericalorange

1,147110

asked Aug 4 at 1:02

numericalorange

1,147110

asked Aug 4 at 1:02

numericalorange

1,147110

1,147110

put on hold as off-topic by Derek Holt, Jyrki Lahtonen, amWhy, José Carlos Santos, John Ma 2 days ago

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Jyrki Lahtonen, amWhy, José Carlos Santos, John Ma

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by Derek Holt, Jyrki Lahtonen, amWhy, José Carlos Santos, John Ma 2 days ago

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Jyrki Lahtonen, amWhy, José Carlos Santos, John Ma

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you men is there nonabelian $G$ and subgroups $H,K$ for which $HK$ is not a subgroup? [Your use of double negative made me ask]
  – coffeemath
  Aug 4 at 1:07
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you men is there nonabelian $G$ and subgroups $H,K$ for which $HK$ is not a subgroup? [Your use of double negative made me ask]
  – coffeemath
  Aug 4 at 1:07
  
  

  
  
  
  

Do you men is there nonabelian $G$ and subgroups $H,K$ for which $HK$ is not a subgroup? [Your use of double negative made me ask]
– coffeemath
Aug 4 at 1:07

Do you men is there nonabelian $G$ and subgroups $H,K$ for which $HK$ is not a subgroup? [Your use of double negative made me ask]
– coffeemath
Aug 4 at 1:07

add a comment | 

3 Answers
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accepted

Consider $G=S_3,H=langle (12)rangle$ and $K=langle (13)rangle$.

By product formula, $$|HK|=fracHcap K=4$$
If $HK$ is a subgroup of $G$, then by Lagrange's Theorem,$|HK|$ divides $|G|$, which is a contradiction.

Hence $HK$ is not a subgroup of $G$.

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answered Aug 4 at 1:10

Alan Wang

3,986930

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Plus one. This is the example I chose, but the reason that the set of products is not a group is much easier: these are $e,(12),(13), (132)$, which do not make up a subgroup, for any number of reasons. Like: the violation of Lagrange, as you said; or the fact that $(13)(12)$ is not in the above list; or that the above set is not one of the easily-enumerable and easily-describable six subgroups of $S_3$.
  – Lubin
  Aug 4 at 2:52
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

In the symmetric group $S_4$, you can take $H = langle (1,2,3)rangle$ and $K=langle (1,4)(2,3)rangle$ (among many other choices). Then it's easy to check that $HK neq KH$, so $HK$ is not a subgroup of $S_4$. (Or, just directly check that $(1,2,4), (1,2,3)in HK$, but $(1,2,4)cdot(1,2,3) = (1,3)(2,4)notin HK$.)

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answered Aug 4 at 1:13

James

6,68721426

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Take $G=S_3,H=left<(1,2)right>$ and $K=left<(2,3)right>$. Then $H$ and $K$ are subgroups of $G($each containing two elements$)$ and $$HK=1,(12),(23),(132)$$a set of size $4$. Therefore, $HK$ is not a subgropu of $G$$($by Lagrange's Theorem$)$ since $4$ does not divide $6$.

share|cite|improve this answer

answered 2 days ago

Key Flex

3,663422

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Consider $G=S_3,H=langle (12)rangle$ and $K=langle (13)rangle$.

By product formula, $$|HK|=fracHcap K=4$$
If $HK$ is a subgroup of $G$, then by Lagrange's Theorem,$|HK|$ divides $|G|$, which is a contradiction.

Hence $HK$ is not a subgroup of $G$.

share|cite|improve this answer

answered Aug 4 at 1:10

Alan Wang

3,986930

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Plus one. This is the example I chose, but the reason that the set of products is not a group is much easier: these are $e,(12),(13), (132)$, which do not make up a subgroup, for any number of reasons. Like: the violation of Lagrange, as you said; or the fact that $(13)(12)$ is not in the above list; or that the above set is not one of the easily-enumerable and easily-describable six subgroups of $S_3$.
  – Lubin
  Aug 4 at 2:52
  
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

Consider $G=S_3,H=langle (12)rangle$ and $K=langle (13)rangle$.

By product formula, $$|HK|=fracHcap K=4$$
If $HK$ is a subgroup of $G$, then by Lagrange's Theorem,$|HK|$ divides $|G|$, which is a contradiction.

Hence $HK$ is not a subgroup of $G$.

share|cite|improve this answer

answered Aug 4 at 1:10

Alan Wang

3,986930

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Plus one. This is the example I chose, but the reason that the set of products is not a group is much easier: these are $e,(12),(13), (132)$, which do not make up a subgroup, for any number of reasons. Like: the violation of Lagrange, as you said; or the fact that $(13)(12)$ is not in the above list; or that the above set is not one of the easily-enumerable and easily-describable six subgroups of $S_3$.
  – Lubin
  Aug 4 at 2:52
  
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Consider $G=S_3,H=langle (12)rangle$ and $K=langle (13)rangle$.

By product formula, $$|HK|=fracHcap K=4$$
If $HK$ is a subgroup of $G$, then by Lagrange's Theorem,$|HK|$ divides $|G|$, which is a contradiction.

Hence $HK$ is not a subgroup of $G$.

share|cite|improve this answer

answered Aug 4 at 1:10

Alan Wang

3,986930

Consider $G=S_3,H=langle (12)rangle$ and $K=langle (13)rangle$.

By product formula, $$|HK|=fracHcap K=4$$
If $HK$ is a subgroup of $G$, then by Lagrange's Theorem,$|HK|$ divides $|G|$, which is a contradiction.

Hence $HK$ is not a subgroup of $G$.

share|cite|improve this answer

answered Aug 4 at 1:10

Alan Wang

3,986930

share|cite|improve this answer

share|cite|improve this answer

answered Aug 4 at 1:10

Alan Wang

3,986930

answered Aug 4 at 1:10

Alan Wang

3,986930

answered Aug 4 at 1:10

Alan Wang

3,986930

3,986930

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Plus one. This is the example I chose, but the reason that the set of products is not a group is much easier: these are $e,(12),(13), (132)$, which do not make up a subgroup, for any number of reasons. Like: the violation of Lagrange, as you said; or the fact that $(13)(12)$ is not in the above list; or that the above set is not one of the easily-enumerable and easily-describable six subgroups of $S_3$.
  – Lubin
  Aug 4 at 2:52
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Plus one. This is the example I chose, but the reason that the set of products is not a group is much easier: these are $e,(12),(13), (132)$, which do not make up a subgroup, for any number of reasons. Like: the violation of Lagrange, as you said; or the fact that $(13)(12)$ is not in the above list; or that the above set is not one of the easily-enumerable and easily-describable six subgroups of $S_3$.
  – Lubin
  Aug 4 at 2:52
  
  

  
  
  
  

1

1

Plus one. This is the example I chose, but the reason that the set of products is not a group is much easier: these are $e,(12),(13), (132)$, which do not make up a subgroup, for any number of reasons. Like: the violation of Lagrange, as you said; or the fact that $(13)(12)$ is not in the above list; or that the above set is not one of the easily-enumerable and easily-describable six subgroups of $S_3$.
– Lubin
Aug 4 at 2:52

Plus one. This is the example I chose, but the reason that the set of products is not a group is much easier: these are $e,(12),(13), (132)$, which do not make up a subgroup, for any number of reasons. Like: the violation of Lagrange, as you said; or the fact that $(13)(12)$ is not in the above list; or that the above set is not one of the easily-enumerable and easily-describable six subgroups of $S_3$.
– Lubin
Aug 4 at 2:52

add a comment | 

up vote
2
down vote

In the symmetric group $S_4$, you can take $H = langle (1,2,3)rangle$ and $K=langle (1,4)(2,3)rangle$ (among many other choices). Then it's easy to check that $HK neq KH$, so $HK$ is not a subgroup of $S_4$. (Or, just directly check that $(1,2,4), (1,2,3)in HK$, but $(1,2,4)cdot(1,2,3) = (1,3)(2,4)notin HK$.)

share|cite|improve this answer

answered Aug 4 at 1:13

James

6,68721426

add a comment | 

up vote
2
down vote

In the symmetric group $S_4$, you can take $H = langle (1,2,3)rangle$ and $K=langle (1,4)(2,3)rangle$ (among many other choices). Then it's easy to check that $HK neq KH$, so $HK$ is not a subgroup of $S_4$. (Or, just directly check that $(1,2,4), (1,2,3)in HK$, but $(1,2,4)cdot(1,2,3) = (1,3)(2,4)notin HK$.)

share|cite|improve this answer

answered Aug 4 at 1:13

James

6,68721426

add a comment | 

up vote
2
down vote

up vote
2
down vote

In the symmetric group $S_4$, you can take $H = langle (1,2,3)rangle$ and $K=langle (1,4)(2,3)rangle$ (among many other choices). Then it's easy to check that $HK neq KH$, so $HK$ is not a subgroup of $S_4$. (Or, just directly check that $(1,2,4), (1,2,3)in HK$, but $(1,2,4)cdot(1,2,3) = (1,3)(2,4)notin HK$.)

share|cite|improve this answer

answered Aug 4 at 1:13

James

6,68721426

In the symmetric group $S_4$, you can take $H = langle (1,2,3)rangle$ and $K=langle (1,4)(2,3)rangle$ (among many other choices). Then it's easy to check that $HK neq KH$, so $HK$ is not a subgroup of $S_4$. (Or, just directly check that $(1,2,4), (1,2,3)in HK$, but $(1,2,4)cdot(1,2,3) = (1,3)(2,4)notin HK$.)

share|cite|improve this answer

answered Aug 4 at 1:13

James

6,68721426

share|cite|improve this answer

share|cite|improve this answer

answered Aug 4 at 1:13

James

6,68721426

answered Aug 4 at 1:13

James

6,68721426

answered Aug 4 at 1:13

James

6,68721426

6,68721426

add a comment | 

add a comment | 

up vote
1
down vote

Take $G=S_3,H=left<(1,2)right>$ and $K=left<(2,3)right>$. Then $H$ and $K$ are subgroups of $G($each containing two elements$)$ and $$HK=1,(12),(23),(132)$$a set of size $4$. Therefore, $HK$ is not a subgropu of $G$$($by Lagrange's Theorem$)$ since $4$ does not divide $6$.

share|cite|improve this answer

answered 2 days ago

Key Flex

3,663422

add a comment | 

up vote
1
down vote

Take $G=S_3,H=left<(1,2)right>$ and $K=left<(2,3)right>$. Then $H$ and $K$ are subgroups of $G($each containing two elements$)$ and $$HK=1,(12),(23),(132)$$a set of size $4$. Therefore, $HK$ is not a subgropu of $G$$($by Lagrange's Theorem$)$ since $4$ does not divide $6$.

share|cite|improve this answer

answered 2 days ago

Key Flex

3,663422

add a comment | 

up vote
1
down vote

up vote
1
down vote

Take $G=S_3,H=left<(1,2)right>$ and $K=left<(2,3)right>$. Then $H$ and $K$ are subgroups of $G($each containing two elements$)$ and $$HK=1,(12),(23),(132)$$a set of size $4$. Therefore, $HK$ is not a subgropu of $G$$($by Lagrange's Theorem$)$ since $4$ does not divide $6$.

share|cite|improve this answer

answered 2 days ago

Key Flex

3,663422

Take $G=S_3,H=left<(1,2)right>$ and $K=left<(2,3)right>$. Then $H$ and $K$ are subgroups of $G($each containing two elements$)$ and $$HK=1,(12),(23),(132)$$a set of size $4$. Therefore, $HK$ is not a subgropu of $G$$($by Lagrange's Theorem$)$ since $4$ does not divide $6$.

share|cite|improve this answer

answered 2 days ago

Key Flex

3,663422

share|cite|improve this answer

share|cite|improve this answer

answered 2 days ago

Key Flex

3,663422

answered 2 days ago

Key Flex

3,663422

answered 2 days ago

Key Flex

3,663422

3,663422

add a comment | 

add a comment |