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Direct product decomposition of the group of complex roots of unity

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1

1

$begingroup$

I'm studying $p$-adic numbers (Robert's "A course in $p$-adic analysis) and, at page 41, the author states that, for every prime $p$, the group $mu$ of all complex roots of unity has a direct product decomposition in terms of $mu_(p)$, the group of roots of unity of order prime to $p$, and $mu_p^infty$, the group of roots of unity having order a power of $p$; explicitly
$$mu=mu_(p)cdotmu_p^infty.$$

I don't understand how I can prove it: if $zin mu$ and $mge2$ (the case $m=1$ is obvious), maybe I can use the exponential form of complex numbers $z=exp(2pi ifrackm)$, but here I stuck.

Any hint is appreciated.
Thank you in advance for your help.

What I found in the web until now it sufficies to observe that $mu$ is abelian torsion and apply the corresponding classification theorem, right?

Note: however, I would like a direct proof.

complex-analysis group-theory complex-numbers abelian-groups roots-of-unity

share|cite|improve this question

edited Mar 21 at 20:53

LBJFS

asked Mar 21 at 19:26

LBJFSLBJFS

360112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you know that classification theorem, then yes
  $endgroup$
  – Max
  Mar 21 at 20:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually I don't know it... I read about it in the web after I posted the question...
  $endgroup$
  – LBJFS
  Mar 21 at 20:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was looking for a "direct" answer, whenever possible
  $endgroup$
  – LBJFS
  Mar 21 at 20:40
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I'm studying $p$-adic numbers (Robert's "A course in $p$-adic analysis) and, at page 41, the author states that, for every prime $p$, the group $mu$ of all complex roots of unity has a direct product decomposition in terms of $mu_(p)$, the group of roots of unity of order prime to $p$, and $mu_p^infty$, the group of roots of unity having order a power of $p$; explicitly
$$mu=mu_(p)cdotmu_p^infty.$$

I don't understand how I can prove it: if $zin mu$ and $mge2$ (the case $m=1$ is obvious), maybe I can use the exponential form of complex numbers $z=exp(2pi ifrackm)$, but here I stuck.

Any hint is appreciated.
Thank you in advance for your help.

What I found in the web until now it sufficies to observe that $mu$ is abelian torsion and apply the corresponding classification theorem, right?

Note: however, I would like a direct proof.

complex-analysis group-theory complex-numbers abelian-groups roots-of-unity

share|cite|improve this question

edited Mar 21 at 20:53

LBJFS

asked Mar 21 at 19:26

LBJFSLBJFS

360112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you know that classification theorem, then yes
  $endgroup$
  – Max
  Mar 21 at 20:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually I don't know it... I read about it in the web after I posted the question...
  $endgroup$
  – LBJFS
  Mar 21 at 20:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was looking for a "direct" answer, whenever possible
  $endgroup$
  – LBJFS
  Mar 21 at 20:40
  

  
  
  
  

add a comment | 

$begingroup$

I'm studying $p$-adic numbers (Robert's "A course in $p$-adic analysis) and, at page 41, the author states that, for every prime $p$, the group $mu$ of all complex roots of unity has a direct product decomposition in terms of $mu_(p)$, the group of roots of unity of order prime to $p$, and $mu_p^infty$, the group of roots of unity having order a power of $p$; explicitly
$$mu=mu_(p)cdotmu_p^infty.$$

I don't understand how I can prove it: if $zin mu$ and $mge2$ (the case $m=1$ is obvious), maybe I can use the exponential form of complex numbers $z=exp(2pi ifrackm)$, but here I stuck.

Any hint is appreciated.
Thank you in advance for your help.

What I found in the web until now it sufficies to observe that $mu$ is abelian torsion and apply the corresponding classification theorem, right?

Note: however, I would like a direct proof.

complex-analysis group-theory complex-numbers abelian-groups roots-of-unity

share|cite|improve this question

edited Mar 21 at 20:53

LBJFS

asked Mar 21 at 19:26

LBJFSLBJFS

360112

$endgroup$

share|cite|improve this question

edited Mar 21 at 20:53

LBJFS

asked Mar 21 at 19:26

LBJFSLBJFS

360112

share|cite|improve this question

edited Mar 21 at 20:53

LBJFS

asked Mar 21 at 19:26

LBJFSLBJFS

360112

asked Mar 21 at 19:26

LBJFSLBJFS

360112

asked Mar 21 at 19:26

LBJFSLBJFS

360112

1 Answer
1

active

oldest

votes

1

$begingroup$

If you want a direct proof, the proof of the classification theorem is not so hard anyways.

Indeed, clearly the two subgroups in question have trivial intersection, and they're both normal since $mu$ is abelian. Moreover if $xin mu$ has order $n=p^alpha m$ with $mland p = 1$, then write $um + vp^alpha = 1$ with Bezout's theorem, so that $x= x^um+vp^alpha = x^umx^vp^alpha$ and you can easily check that $x^um$ has order $p^alpha$ and $x^vp^alpha$ has order $m$, that is, order prime with $p$.

The decomposition follows.

share|cite|improve this answer

answered Mar 21 at 20:57

MaxMax

16k11144

$endgroup$

add a comment | 

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1

$begingroup$

If you want a direct proof, the proof of the classification theorem is not so hard anyways.

Indeed, clearly the two subgroups in question have trivial intersection, and they're both normal since $mu$ is abelian. Moreover if $xin mu$ has order $n=p^alpha m$ with $mland p = 1$, then write $um + vp^alpha = 1$ with Bezout's theorem, so that $x= x^um+vp^alpha = x^umx^vp^alpha$ and you can easily check that $x^um$ has order $p^alpha$ and $x^vp^alpha$ has order $m$, that is, order prime with $p$.

The decomposition follows.

share|cite|improve this answer

answered Mar 21 at 20:57

MaxMax

16k11144

$endgroup$

add a comment | 

1

$begingroup$

If you want a direct proof, the proof of the classification theorem is not so hard anyways.

Indeed, clearly the two subgroups in question have trivial intersection, and they're both normal since $mu$ is abelian. Moreover if $xin mu$ has order $n=p^alpha m$ with $mland p = 1$, then write $um + vp^alpha = 1$ with Bezout's theorem, so that $x= x^um+vp^alpha = x^umx^vp^alpha$ and you can easily check that $x^um$ has order $p^alpha$ and $x^vp^alpha$ has order $m$, that is, order prime with $p$.

The decomposition follows.

share|cite|improve this answer

answered Mar 21 at 20:57

MaxMax

16k11144

$endgroup$

add a comment | 

$begingroup$

If you want a direct proof, the proof of the classification theorem is not so hard anyways.

Indeed, clearly the two subgroups in question have trivial intersection, and they're both normal since $mu$ is abelian. Moreover if $xin mu$ has order $n=p^alpha m$ with $mland p = 1$, then write $um + vp^alpha = 1$ with Bezout's theorem, so that $x= x^um+vp^alpha = x^umx^vp^alpha$ and you can easily check that $x^um$ has order $p^alpha$ and $x^vp^alpha$ has order $m$, that is, order prime with $p$.

The decomposition follows.

share|cite|improve this answer

answered Mar 21 at 20:57

MaxMax

16k11144

$endgroup$

share|cite|improve this answer

answered Mar 21 at 20:57

MaxMax

16k11144

answered Mar 21 at 20:57

MaxMax

16k11144

answered Mar 21 at 20:57

MaxMax

16k11144