Order of elements of the Prüfer groups $mathbb{Z}(p^{infty})$ [duplicate]

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  Characterising subgroups of Prüfer $p$-groups.
  
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Let $mathbb{Z}(p^{infty})$ be defined by

$mathbb{Z}(p^{infty}) = { overline{a/b} in mathbb{Q}/ mathbb{Z} / a,b in mathbb{Z}, b=p^i$ $ with$ $ i in mathbb{N} }$, I wish show that any element in $mathbb{Z}(p^{infty})$ has order $p^n$ with $n in mathbb{N}$.

i try several ways but I have not been successful,

some help ??

thank you

abstract-algebra p-groups

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edited Jan 29 at 9:35

YuiTo Cheng

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asked Jan 29 at 8:53

ahren.bazahren.baz

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Jan 29 at 9:01

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$begingroup$

This question already has an answer here:

• 
  Characterising subgroups of Prüfer $p$-groups.
  
  2 answers
  
  

Let $mathbb{Z}(p^{infty})$ be defined by

$mathbb{Z}(p^{infty}) = { overline{a/b} in mathbb{Q}/ mathbb{Z} / a,b in mathbb{Z}, b=p^i$ $ with$ $ i in mathbb{N} }$, I wish show that any element in $mathbb{Z}(p^{infty})$ has order $p^n$ with $n in mathbb{N}$.

i try several ways but I have not been successful,

some help ??

thank you

abstract-algebra p-groups

share|cite|improve this question

edited Jan 29 at 9:35

YuiTo Cheng

2,1862937

asked Jan 29 at 8:53

ahren.bazahren.baz

263

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marked as duplicate by Dietrich Burde abstract-algebra
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$begingroup$

This question already has an answer here:

• 
  Characterising subgroups of Prüfer $p$-groups.
  
  2 answers
  
  

Let $mathbb{Z}(p^{infty})$ be defined by

$mathbb{Z}(p^{infty}) = { overline{a/b} in mathbb{Q}/ mathbb{Z} / a,b in mathbb{Z}, b=p^i$ $ with$ $ i in mathbb{N} }$, I wish show that any element in $mathbb{Z}(p^{infty})$ has order $p^n$ with $n in mathbb{N}$.

i try several ways but I have not been successful,

some help ??

thank you

abstract-algebra p-groups

share|cite|improve this question

edited Jan 29 at 9:35

YuiTo Cheng

2,1862937

asked Jan 29 at 8:53

ahren.bazahren.baz

263

$endgroup$

This question already has an answer here:

• 
  Characterising subgroups of Prüfer $p$-groups.
  
  2 answers
  
  

Let $mathbb{Z}(p^{infty})$ be defined by

$mathbb{Z}(p^{infty}) = { overline{a/b} in mathbb{Q}/ mathbb{Z} / a,b in mathbb{Z}, b=p^i$ $ with$ $ i in mathbb{N} }$, I wish show that any element in $mathbb{Z}(p^{infty})$ has order $p^n$ with $n in mathbb{N}$.

i try several ways but I have not been successful,

some help ??

thank you

This question already has an answer here:

• 
  Characterising subgroups of Prüfer $p$-groups.
  
  2 answers
  
  

abstract-algebra p-groups

abstract-algebra p-groups

share|cite|improve this question

edited Jan 29 at 9:35

YuiTo Cheng

2,1862937

asked Jan 29 at 8:53

ahren.bazahren.baz

263

share|cite|improve this question

edited Jan 29 at 9:35

YuiTo Cheng

2,1862937

asked Jan 29 at 8:53

ahren.bazahren.baz

263

share|cite|improve this question

share|cite|improve this question

edited Jan 29 at 9:35

YuiTo Cheng

2,1862937

edited Jan 29 at 9:35

YuiTo Cheng

2,1862937

edited Jan 29 at 9:35

YuiTo Cheng

2,1862937

2,1862937

asked Jan 29 at 8:53

ahren.bazahren.baz

263

asked Jan 29 at 8:53

ahren.bazahren.baz

263

asked Jan 29 at 8:53

ahren.bazahren.baz

263

263

marked as duplicate by Dietrich Burde abstract-algebra
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Jan 29 at 9:01

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1 Answer
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I cite from this duplicate:

Every element of ${bf Z}_{p^infty}$ is represented by some $a/p^n$, where $0leq a<p^n$ (this is mostly immediate by the definition). I will simply identify this number with its class modulo ${bf Z}$. Furthermore, clearly the order of $a/p^n$ is $p^n$ whenever $a$ is not divisible by $p$.

A further duplicate is here:

Characterising subgroups of Prüfer $p$-groups.

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answered Jan 29 at 8:59

Dietrich BurdeDietrich Burde

81.6k648106

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

$begingroup$

I cite from this duplicate:

Every element of ${bf Z}_{p^infty}$ is represented by some $a/p^n$, where $0leq a<p^n$ (this is mostly immediate by the definition). I will simply identify this number with its class modulo ${bf Z}$. Furthermore, clearly the order of $a/p^n$ is $p^n$ whenever $a$ is not divisible by $p$.

A further duplicate is here:

Characterising subgroups of Prüfer $p$-groups.

share|cite|improve this answer

answered Jan 29 at 8:59

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

add a comment | 

0

$begingroup$

I cite from this duplicate:

Every element of ${bf Z}_{p^infty}$ is represented by some $a/p^n$, where $0leq a<p^n$ (this is mostly immediate by the definition). I will simply identify this number with its class modulo ${bf Z}$. Furthermore, clearly the order of $a/p^n$ is $p^n$ whenever $a$ is not divisible by $p$.

A further duplicate is here:

Characterising subgroups of Prüfer $p$-groups.

share|cite|improve this answer

answered Jan 29 at 8:59

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

add a comment | 

0

0

0

$begingroup$

I cite from this duplicate:

Every element of ${bf Z}_{p^infty}$ is represented by some $a/p^n$, where $0leq a<p^n$ (this is mostly immediate by the definition). I will simply identify this number with its class modulo ${bf Z}$. Furthermore, clearly the order of $a/p^n$ is $p^n$ whenever $a$ is not divisible by $p$.

A further duplicate is here:

Characterising subgroups of Prüfer $p$-groups.

share|cite|improve this answer

answered Jan 29 at 8:59

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

I cite from this duplicate:

Every element of ${bf Z}_{p^infty}$ is represented by some $a/p^n$, where $0leq a<p^n$ (this is mostly immediate by the definition). I will simply identify this number with its class modulo ${bf Z}$. Furthermore, clearly the order of $a/p^n$ is $p^n$ whenever $a$ is not divisible by $p$.

A further duplicate is here:

Characterising subgroups of Prüfer $p$-groups.

share|cite|improve this answer

answered Jan 29 at 8:59

Dietrich BurdeDietrich Burde

81.6k648106

share|cite|improve this answer

share|cite|improve this answer

answered Jan 29 at 8:59

Dietrich BurdeDietrich Burde

81.6k648106

answered Jan 29 at 8:59

Dietrich BurdeDietrich Burde

81.6k648106

answered Jan 29 at 8:59

Dietrich BurdeDietrich Burde

81.6k648106

81.6k648106

add a comment | 

add a comment | 

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