How to show that $mathbbZ_12 $ is isomorphic to a subgroup of $S_7$?

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How to show that $mathbbZ_12$ is isomorphic to a subgroup of $S_7$?

My attempt: Using Cayley's theorem one can conclude $mathbbZ_12$ is isomorphic to a subgroup of $S_12$.

Or, if I use Generalised Cayley's theorem I can show that there is a homomorphism from $mathbbZ_12rightarrow S_mathbbZ_12/H$ where $H$ is a subgroup of order $3,2^2$, therefore we have group homomorphism from $mathbbZ_12$ to $S_3$ or $S_4$. But these maps have non-trivial kernel namely $H$ itself.

Therefore, I have not been able to conclude the required statement.

Any help is appreciated.

group-theory permutation-cycles

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asked Jul 30 at 9:03

Babai

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up vote
1
down vote

favorite

How to show that $mathbbZ_12$ is isomorphic to a subgroup of $S_7$?

My attempt: Using Cayley's theorem one can conclude $mathbbZ_12$ is isomorphic to a subgroup of $S_12$.

Or, if I use Generalised Cayley's theorem I can show that there is a homomorphism from $mathbbZ_12rightarrow S_mathbbZ_12/H$ where $H$ is a subgroup of order $3,2^2$, therefore we have group homomorphism from $mathbbZ_12$ to $S_3$ or $S_4$. But these maps have non-trivial kernel namely $H$ itself.

Therefore, I have not been able to conclude the required statement.

Any help is appreciated.

group-theory permutation-cycles

share|cite|improve this question

asked Jul 30 at 9:03

Babai

2,50021539

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

How to show that $mathbbZ_12$ is isomorphic to a subgroup of $S_7$?

My attempt: Using Cayley's theorem one can conclude $mathbbZ_12$ is isomorphic to a subgroup of $S_12$.

Or, if I use Generalised Cayley's theorem I can show that there is a homomorphism from $mathbbZ_12rightarrow S_mathbbZ_12/H$ where $H$ is a subgroup of order $3,2^2$, therefore we have group homomorphism from $mathbbZ_12$ to $S_3$ or $S_4$. But these maps have non-trivial kernel namely $H$ itself.

Therefore, I have not been able to conclude the required statement.

Any help is appreciated.

group-theory permutation-cycles

share|cite|improve this question

asked Jul 30 at 9:03

Babai

2,50021539

How to show that $mathbbZ_12$ is isomorphic to a subgroup of $S_7$?

My attempt: Using Cayley's theorem one can conclude $mathbbZ_12$ is isomorphic to a subgroup of $S_12$.

Or, if I use Generalised Cayley's theorem I can show that there is a homomorphism from $mathbbZ_12rightarrow S_mathbbZ_12/H$ where $H$ is a subgroup of order $3,2^2$, therefore we have group homomorphism from $mathbbZ_12$ to $S_3$ or $S_4$. But these maps have non-trivial kernel namely $H$ itself.

Therefore, I have not been able to conclude the required statement.

Any help is appreciated.

group-theory permutation-cycles

share|cite|improve this question

asked Jul 30 at 9:03

Babai

2,50021539

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 9:03

Babai

2,50021539

asked Jul 30 at 9:03

Babai

2,50021539

asked Jul 30 at 9:03

Babai

2,50021539

2,50021539

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
6
down vote



accepted

Hint: look at $(1 2 3)(4 5 6 7)$. What is its order?

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answered Jul 30 at 9:05

Nicky Hekster

26.9k53052

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The order is 12. Hence, the subgroup generated by is a cyclic subgroup of order 12. And also, any cyclic subgroups of the same order are isomorphic, hence we are done. Thank you.
  – Babai
  Jul 30 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly. Well done!
  – Nicky Hekster
  Jul 30 at 9:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Babal, if you deem my answer as being the right one please tick it as such. Thanks!
  – Nicky Hekster
  Jul 30 at 9:57
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

In this case, as Nicky Hekster pointed out in the top answer, you know the general structure of elements in $S_7$ so finding one of order $12$ is pretty easy.

I want to add that a more general procedure to look for a copy of $mathbbZ_12$ in a group $G$ is to consider that $mathbbZ_12 = mathbbZ_4 times mathbbZ_3$. So you can look for an element of order $3$ (computing the Sylow $3$-subgroup) and then seeing if the centraliser of this element has an element of order $4$ (or, the dual procedure, looking for a copy of $mathbbZ_4$ in the Sylow $2$-subgroup of $G$ and then computing the order of its centraliser, if $3$ divides it you are done).

Of course, the difficulty in this lies in the fact that computing Sylow subgroups or centralisers might not be easy, but in some cases it is. (First example I can think of, if $G=S_4 times S_4$ you can take $((1234), 1)$ as the element of order $4$ and it is immediate that $(1,(123))$ is contained in its centraliser).

share|cite|improve this answer

answered Jul 30 at 10:51

AnalysisStudent0414

4,206828

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
6
down vote



accepted

Hint: look at $(1 2 3)(4 5 6 7)$. What is its order?

share|cite|improve this answer

answered Jul 30 at 9:05

Nicky Hekster

26.9k53052

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The order is 12. Hence, the subgroup generated by is a cyclic subgroup of order 12. And also, any cyclic subgroups of the same order are isomorphic, hence we are done. Thank you.
  – Babai
  Jul 30 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly. Well done!
  – Nicky Hekster
  Jul 30 at 9:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Babal, if you deem my answer as being the right one please tick it as such. Thanks!
  – Nicky Hekster
  Jul 30 at 9:57
  
  

  
  
  
  

add a comment | 

up vote
6
down vote



accepted

Hint: look at $(1 2 3)(4 5 6 7)$. What is its order?

share|cite|improve this answer

answered Jul 30 at 9:05

Nicky Hekster

26.9k53052

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The order is 12. Hence, the subgroup generated by is a cyclic subgroup of order 12. And also, any cyclic subgroups of the same order are isomorphic, hence we are done. Thank you.
  – Babai
  Jul 30 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly. Well done!
  – Nicky Hekster
  Jul 30 at 9:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Babal, if you deem my answer as being the right one please tick it as such. Thanks!
  – Nicky Hekster
  Jul 30 at 9:57
  
  

  
  
  
  

add a comment | 

up vote
6
down vote



accepted

up vote
6
down vote



accepted

Hint: look at $(1 2 3)(4 5 6 7)$. What is its order?

share|cite|improve this answer

answered Jul 30 at 9:05

Nicky Hekster

26.9k53052

Hint: look at $(1 2 3)(4 5 6 7)$. What is its order?

share|cite|improve this answer

answered Jul 30 at 9:05

Nicky Hekster

26.9k53052

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 9:05

Nicky Hekster

26.9k53052

answered Jul 30 at 9:05

Nicky Hekster

26.9k53052

answered Jul 30 at 9:05

Nicky Hekster

26.9k53052

26.9k53052

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The order is 12. Hence, the subgroup generated by is a cyclic subgroup of order 12. And also, any cyclic subgroups of the same order are isomorphic, hence we are done. Thank you.
  – Babai
  Jul 30 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly. Well done!
  – Nicky Hekster
  Jul 30 at 9:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Babal, if you deem my answer as being the right one please tick it as such. Thanks!
  – Nicky Hekster
  Jul 30 at 9:57
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The order is 12. Hence, the subgroup generated by is a cyclic subgroup of order 12. And also, any cyclic subgroups of the same order are isomorphic, hence we are done. Thank you.
  – Babai
  Jul 30 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes exactly. Well done!
  – Nicky Hekster
  Jul 30 at 9:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Babal, if you deem my answer as being the right one please tick it as such. Thanks!
  – Nicky Hekster
  Jul 30 at 9:57
  
  

  
  
  
  

The order is 12. Hence, the subgroup generated by is a cyclic subgroup of order 12. And also, any cyclic subgroups of the same order are isomorphic, hence we are done. Thank you.
– Babai
Jul 30 at 9:09

The order is 12. Hence, the subgroup generated by is a cyclic subgroup of order 12. And also, any cyclic subgroups of the same order are isomorphic, hence we are done. Thank you.
– Babai
Jul 30 at 9:09

Yes exactly. Well done!
– Nicky Hekster
Jul 30 at 9:44

Yes exactly. Well done!
– Nicky Hekster
Jul 30 at 9:44

Babal, if you deem my answer as being the right one please tick it as such. Thanks!
– Nicky Hekster
Jul 30 at 9:57

Babal, if you deem my answer as being the right one please tick it as such. Thanks!
– Nicky Hekster
Jul 30 at 9:57

add a comment | 

up vote
0
down vote

In this case, as Nicky Hekster pointed out in the top answer, you know the general structure of elements in $S_7$ so finding one of order $12$ is pretty easy.

I want to add that a more general procedure to look for a copy of $mathbbZ_12$ in a group $G$ is to consider that $mathbbZ_12 = mathbbZ_4 times mathbbZ_3$. So you can look for an element of order $3$ (computing the Sylow $3$-subgroup) and then seeing if the centraliser of this element has an element of order $4$ (or, the dual procedure, looking for a copy of $mathbbZ_4$ in the Sylow $2$-subgroup of $G$ and then computing the order of its centraliser, if $3$ divides it you are done).

Of course, the difficulty in this lies in the fact that computing Sylow subgroups or centralisers might not be easy, but in some cases it is. (First example I can think of, if $G=S_4 times S_4$ you can take $((1234), 1)$ as the element of order $4$ and it is immediate that $(1,(123))$ is contained in its centraliser).

share|cite|improve this answer

answered Jul 30 at 10:51

AnalysisStudent0414

4,206828

add a comment | 

up vote
0
down vote

In this case, as Nicky Hekster pointed out in the top answer, you know the general structure of elements in $S_7$ so finding one of order $12$ is pretty easy.

I want to add that a more general procedure to look for a copy of $mathbbZ_12$ in a group $G$ is to consider that $mathbbZ_12 = mathbbZ_4 times mathbbZ_3$. So you can look for an element of order $3$ (computing the Sylow $3$-subgroup) and then seeing if the centraliser of this element has an element of order $4$ (or, the dual procedure, looking for a copy of $mathbbZ_4$ in the Sylow $2$-subgroup of $G$ and then computing the order of its centraliser, if $3$ divides it you are done).

Of course, the difficulty in this lies in the fact that computing Sylow subgroups or centralisers might not be easy, but in some cases it is. (First example I can think of, if $G=S_4 times S_4$ you can take $((1234), 1)$ as the element of order $4$ and it is immediate that $(1,(123))$ is contained in its centraliser).

share|cite|improve this answer

answered Jul 30 at 10:51

AnalysisStudent0414

4,206828

add a comment | 

up vote
0
down vote

up vote
0
down vote

In this case, as Nicky Hekster pointed out in the top answer, you know the general structure of elements in $S_7$ so finding one of order $12$ is pretty easy.

I want to add that a more general procedure to look for a copy of $mathbbZ_12$ in a group $G$ is to consider that $mathbbZ_12 = mathbbZ_4 times mathbbZ_3$. So you can look for an element of order $3$ (computing the Sylow $3$-subgroup) and then seeing if the centraliser of this element has an element of order $4$ (or, the dual procedure, looking for a copy of $mathbbZ_4$ in the Sylow $2$-subgroup of $G$ and then computing the order of its centraliser, if $3$ divides it you are done).

Of course, the difficulty in this lies in the fact that computing Sylow subgroups or centralisers might not be easy, but in some cases it is. (First example I can think of, if $G=S_4 times S_4$ you can take $((1234), 1)$ as the element of order $4$ and it is immediate that $(1,(123))$ is contained in its centraliser).

share|cite|improve this answer

answered Jul 30 at 10:51

AnalysisStudent0414

4,206828

In this case, as Nicky Hekster pointed out in the top answer, you know the general structure of elements in $S_7$ so finding one of order $12$ is pretty easy.

I want to add that a more general procedure to look for a copy of $mathbbZ_12$ in a group $G$ is to consider that $mathbbZ_12 = mathbbZ_4 times mathbbZ_3$. So you can look for an element of order $3$ (computing the Sylow $3$-subgroup) and then seeing if the centraliser of this element has an element of order $4$ (or, the dual procedure, looking for a copy of $mathbbZ_4$ in the Sylow $2$-subgroup of $G$ and then computing the order of its centraliser, if $3$ divides it you are done).

Of course, the difficulty in this lies in the fact that computing Sylow subgroups or centralisers might not be easy, but in some cases it is. (First example I can think of, if $G=S_4 times S_4$ you can take $((1234), 1)$ as the element of order $4$ and it is immediate that $(1,(123))$ is contained in its centraliser).

share|cite|improve this answer

answered Jul 30 at 10:51

AnalysisStudent0414

4,206828

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 10:51

AnalysisStudent0414

4,206828

answered Jul 30 at 10:51

AnalysisStudent0414

4,206828

answered Jul 30 at 10:51

AnalysisStudent0414

4,206828

4,206828

add a comment | 

add a comment | 

 

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