Free module over a group transfers to a free module over a subgroup

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Let $H$ be a subgroup of $G$.

1- Why is $mathbbZG$ a free module over $mathbbZH$?

2- Why does every free $mathbbZG$-module is also free $mathbbZH$-module?

group-theory modules free-modules

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asked Jul 30 at 13:15

M.Ramana

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Let $H$ be a subgroup of $G$.

1- Why is $mathbbZG$ a free module over $mathbbZH$?

2- Why does every free $mathbbZG$-module is also free $mathbbZH$-module?

group-theory modules free-modules

share|cite|improve this question

asked Jul 30 at 13:15

M.Ramana

3959

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $H$ be a subgroup of $G$.

1- Why is $mathbbZG$ a free module over $mathbbZH$?

2- Why does every free $mathbbZG$-module is also free $mathbbZH$-module?

group-theory modules free-modules

share|cite|improve this question

asked Jul 30 at 13:15

M.Ramana

3959

Let $H$ be a subgroup of $G$.

1- Why is $mathbbZG$ a free module over $mathbbZH$?

2- Why does every free $mathbbZG$-module is also free $mathbbZH$-module?

group-theory modules free-modules

share|cite|improve this question

asked Jul 30 at 13:15

M.Ramana

3959

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 13:15

M.Ramana

3959

asked Jul 30 at 13:15

M.Ramana

3959

asked Jul 30 at 13:15

M.Ramana

3959

3959

add a comment | 

add a comment | 

1 Answer
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Note that you have $G = bigcup_g in S Hg$ with $S$ being a set of representatives for the right cosets of $G$ with respect to $H$. Now, we have $$mathbbZ G = bigoplus_g in S mathbbZ H g$$ and this is a decomposition of $mathbbZ H$-modules. Note that $mathbbZ H g cong mathbbZ H$ as $mathbbZ H$-modules via multiplication with $g^-1$ from the right. It follows that $mathbbZG$ is the direct sum of copies of $mathbbZH$ and thus $mathbbZG$ is free as a $mathbbZH$-module.

For the second part note that every free $mathbbZG$-module is a direct sum of copis of $mathbbZG$ so the same argument applies.

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answered Jul 30 at 13:27

Matthias Klupsch

6,0341127

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you so much. I understood well.
  – M.Ramana
  Jul 31 at 14:07
  
  

  
  
  
  

add a comment | 

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Note that you have $G = bigcup_g in S Hg$ with $S$ being a set of representatives for the right cosets of $G$ with respect to $H$. Now, we have $$mathbbZ G = bigoplus_g in S mathbbZ H g$$ and this is a decomposition of $mathbbZ H$-modules. Note that $mathbbZ H g cong mathbbZ H$ as $mathbbZ H$-modules via multiplication with $g^-1$ from the right. It follows that $mathbbZG$ is the direct sum of copies of $mathbbZH$ and thus $mathbbZG$ is free as a $mathbbZH$-module.

For the second part note that every free $mathbbZG$-module is a direct sum of copis of $mathbbZG$ so the same argument applies.

share|cite|improve this answer

answered Jul 30 at 13:27

Matthias Klupsch

6,0341127

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you so much. I understood well.
  – M.Ramana
  Jul 31 at 14:07
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Note that you have $G = bigcup_g in S Hg$ with $S$ being a set of representatives for the right cosets of $G$ with respect to $H$. Now, we have $$mathbbZ G = bigoplus_g in S mathbbZ H g$$ and this is a decomposition of $mathbbZ H$-modules. Note that $mathbbZ H g cong mathbbZ H$ as $mathbbZ H$-modules via multiplication with $g^-1$ from the right. It follows that $mathbbZG$ is the direct sum of copies of $mathbbZH$ and thus $mathbbZG$ is free as a $mathbbZH$-module.

For the second part note that every free $mathbbZG$-module is a direct sum of copis of $mathbbZG$ so the same argument applies.

share|cite|improve this answer

answered Jul 30 at 13:27

Matthias Klupsch

6,0341127

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you so much. I understood well.
  – M.Ramana
  Jul 31 at 14:07
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Note that you have $G = bigcup_g in S Hg$ with $S$ being a set of representatives for the right cosets of $G$ with respect to $H$. Now, we have $$mathbbZ G = bigoplus_g in S mathbbZ H g$$ and this is a decomposition of $mathbbZ H$-modules. Note that $mathbbZ H g cong mathbbZ H$ as $mathbbZ H$-modules via multiplication with $g^-1$ from the right. It follows that $mathbbZG$ is the direct sum of copies of $mathbbZH$ and thus $mathbbZG$ is free as a $mathbbZH$-module.

For the second part note that every free $mathbbZG$-module is a direct sum of copis of $mathbbZG$ so the same argument applies.

share|cite|improve this answer

answered Jul 30 at 13:27

Matthias Klupsch

6,0341127

Note that you have $G = bigcup_g in S Hg$ with $S$ being a set of representatives for the right cosets of $G$ with respect to $H$. Now, we have $$mathbbZ G = bigoplus_g in S mathbbZ H g$$ and this is a decomposition of $mathbbZ H$-modules. Note that $mathbbZ H g cong mathbbZ H$ as $mathbbZ H$-modules via multiplication with $g^-1$ from the right. It follows that $mathbbZG$ is the direct sum of copies of $mathbbZH$ and thus $mathbbZG$ is free as a $mathbbZH$-module.

For the second part note that every free $mathbbZG$-module is a direct sum of copis of $mathbbZG$ so the same argument applies.

share|cite|improve this answer

answered Jul 30 at 13:27

Matthias Klupsch

6,0341127

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 13:27

Matthias Klupsch

6,0341127

answered Jul 30 at 13:27

Matthias Klupsch

6,0341127

answered Jul 30 at 13:27

Matthias Klupsch

6,0341127

6,0341127

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you so much. I understood well.
  – M.Ramana
  Jul 31 at 14:07
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you so much. I understood well.
  – M.Ramana
  Jul 31 at 14:07
  
  

  
  
  
  

Thank you so much. I understood well.
– M.Ramana
Jul 31 at 14:07

Thank you so much. I understood well.
– M.Ramana
Jul 31 at 14:07

add a comment | 

 

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