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Let $G$ be a cyclic group of order $12$ generated by $a$ and $H$ be a subgroup of $G$ generated by $a^4$. Find the distinct left cosets of $H$ in $G$.

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0

$begingroup$

Let $G$ be a cyclic group of order $12$ generated by $a$ and $H$ be a subgroup of $G$ generated by $a^4$. Show that the distinct left cosets of $H$ in $G$ are $H, aH, a^2H, a^3H$. Verify that $Hcup a^2H$ is also a subgroup of $G$.

Given $o(G)=o(<a>)=12$ then $a^{12}=e$, $H=<a^4>={e, a^4, a^8}$ therefore, $o(H)=o(<a^4>)=3$. Therefore, number of distinct left cosets of $H$ in $G$ $=[G:H]=frac{o(G)}{o(H)}=12/3=4$.

Clearly, $G={e, a, a^2, a^3, a^4, cdots a^{11}}$. Then the left cosets are $a^rH$, $~~0leq rleq 11$.

The problem can be solved by exploring all 12 cases of $a^rH$, $~~0leq rleq 11$ and then choosing $4$ distinct left cosets will give the desired result.

I have three questions:

1. Is there are any error in the above answer? If there is any modification needed, however small, please feel free to suggest.




2. Exploring all 12 cases of $a^rH$, $~~0leq rleq 11$ and choosing desired cosets is laborious work. Please help me to get the desired cosets so that one can be convinced with the answer.




3. What will be the solution of the last verification part.

abstract-algebra group-theory finite-groups

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asked Sep 27 '15 at 14:36

rama_ranrama_ran

300313

$endgroup$

add a comment | 

0

$begingroup$

Let $G$ be a cyclic group of order $12$ generated by $a$ and $H$ be a subgroup of $G$ generated by $a^4$. Show that the distinct left cosets of $H$ in $G$ are $H, aH, a^2H, a^3H$. Verify that $Hcup a^2H$ is also a subgroup of $G$.

Given $o(G)=o(<a>)=12$ then $a^{12}=e$, $H=<a^4>={e, a^4, a^8}$ therefore, $o(H)=o(<a^4>)=3$. Therefore, number of distinct left cosets of $H$ in $G$ $=[G:H]=frac{o(G)}{o(H)}=12/3=4$.

Clearly, $G={e, a, a^2, a^3, a^4, cdots a^{11}}$. Then the left cosets are $a^rH$, $~~0leq rleq 11$.

The problem can be solved by exploring all 12 cases of $a^rH$, $~~0leq rleq 11$ and then choosing $4$ distinct left cosets will give the desired result.

I have three questions:

1. Is there are any error in the above answer? If there is any modification needed, however small, please feel free to suggest.




2. Exploring all 12 cases of $a^rH$, $~~0leq rleq 11$ and choosing desired cosets is laborious work. Please help me to get the desired cosets so that one can be convinced with the answer.




3. What will be the solution of the last verification part.

abstract-algebra group-theory finite-groups

share|cite|improve this question

asked Sep 27 '15 at 14:36

rama_ranrama_ran

300313

$endgroup$

add a comment | 

$begingroup$

Let $G$ be a cyclic group of order $12$ generated by $a$ and $H$ be a subgroup of $G$ generated by $a^4$. Show that the distinct left cosets of $H$ in $G$ are $H, aH, a^2H, a^3H$. Verify that $Hcup a^2H$ is also a subgroup of $G$.

Given $o(G)=o(<a>)=12$ then $a^{12}=e$, $H=<a^4>={e, a^4, a^8}$ therefore, $o(H)=o(<a^4>)=3$. Therefore, number of distinct left cosets of $H$ in $G$ $=[G:H]=frac{o(G)}{o(H)}=12/3=4$.

Clearly, $G={e, a, a^2, a^3, a^4, cdots a^{11}}$. Then the left cosets are $a^rH$, $~~0leq rleq 11$.

The problem can be solved by exploring all 12 cases of $a^rH$, $~~0leq rleq 11$ and then choosing $4$ distinct left cosets will give the desired result.

I have three questions:

1. Is there are any error in the above answer? If there is any modification needed, however small, please feel free to suggest.




2. Exploring all 12 cases of $a^rH$, $~~0leq rleq 11$ and choosing desired cosets is laborious work. Please help me to get the desired cosets so that one can be convinced with the answer.




3. What will be the solution of the last verification part.

abstract-algebra group-theory finite-groups

share|cite|improve this question

asked Sep 27 '15 at 14:36

rama_ranrama_ran

300313

$endgroup$

share|cite|improve this question

asked Sep 27 '15 at 14:36

rama_ranrama_ran

300313

share|cite|improve this question

asked Sep 27 '15 at 14:36

rama_ranrama_ran

300313

asked Sep 27 '15 at 14:36

rama_ranrama_ran

300313

asked Sep 27 '15 at 14:36

rama_ranrama_ran

300313

1 Answer
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1

$begingroup$

1. The biggest problem as I see it is that you use <a> as angle brackets, rather than langle a rangle: $langle a rangle$. No real mathematical problems.


2. Saying something like "We note that $a^nH=a^{n+4}H$, so $H=a^4H=a^8H$, and so on." would give you a lot less checking. You would only have to check that $H, aH, a^2H$ and $a^3H$ are actually distinct.


3. You verify that $$Hcup a^2H={e, a^4, a^8, a^2, a^6, a^{10}}$$ is a subgroup, the same way you would in any other case. That means checking that it is closed under the group operation, that the identity element is there, and that every element in the set has its inverse in there as well.

share|cite|improve this answer

answered Sep 27 '15 at 14:49

ArthurArthur

118k7118201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the answer. Please explore $a^nH=a^{n+4}H$ and how to check cosets are distinct.
  $endgroup$
  – rama_ran
  Sep 27 '15 at 15:13
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

1. The biggest problem as I see it is that you use <a> as angle brackets, rather than langle a rangle: $langle a rangle$. No real mathematical problems.


2. Saying something like "We note that $a^nH=a^{n+4}H$, so $H=a^4H=a^8H$, and so on." would give you a lot less checking. You would only have to check that $H, aH, a^2H$ and $a^3H$ are actually distinct.


3. You verify that $$Hcup a^2H={e, a^4, a^8, a^2, a^6, a^{10}}$$ is a subgroup, the same way you would in any other case. That means checking that it is closed under the group operation, that the identity element is there, and that every element in the set has its inverse in there as well.

share|cite|improve this answer

answered Sep 27 '15 at 14:49

ArthurArthur

118k7118201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the answer. Please explore $a^nH=a^{n+4}H$ and how to check cosets are distinct.
  $endgroup$
  – rama_ran
  Sep 27 '15 at 15:13
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

1. The biggest problem as I see it is that you use <a> as angle brackets, rather than langle a rangle: $langle a rangle$. No real mathematical problems.


2. Saying something like "We note that $a^nH=a^{n+4}H$, so $H=a^4H=a^8H$, and so on." would give you a lot less checking. You would only have to check that $H, aH, a^2H$ and $a^3H$ are actually distinct.


3. You verify that $$Hcup a^2H={e, a^4, a^8, a^2, a^6, a^{10}}$$ is a subgroup, the same way you would in any other case. That means checking that it is closed under the group operation, that the identity element is there, and that every element in the set has its inverse in there as well.

share|cite|improve this answer

answered Sep 27 '15 at 14:49

ArthurArthur

118k7118201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the answer. Please explore $a^nH=a^{n+4}H$ and how to check cosets are distinct.
  $endgroup$
  – rama_ran
  Sep 27 '15 at 15:13
  
  
  

  
  
  
  

add a comment | 

$begingroup$

1. The biggest problem as I see it is that you use <a> as angle brackets, rather than langle a rangle: $langle a rangle$. No real mathematical problems.


2. Saying something like "We note that $a^nH=a^{n+4}H$, so $H=a^4H=a^8H$, and so on." would give you a lot less checking. You would only have to check that $H, aH, a^2H$ and $a^3H$ are actually distinct.


3. You verify that $$Hcup a^2H={e, a^4, a^8, a^2, a^6, a^{10}}$$ is a subgroup, the same way you would in any other case. That means checking that it is closed under the group operation, that the identity element is there, and that every element in the set has its inverse in there as well.

share|cite|improve this answer

answered Sep 27 '15 at 14:49

ArthurArthur

118k7118201

$endgroup$

share|cite|improve this answer

answered Sep 27 '15 at 14:49

ArthurArthur

118k7118201

answered Sep 27 '15 at 14:49

ArthurArthur

118k7118201

answered Sep 27 '15 at 14:49

ArthurArthur

118k7118201