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Show that $ninmathbb{N}$ is square-free

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$begingroup$

Show that $ninmathbb{N}$ is square-free if and only if there is a subset different from the null set, $Lsubseteqmathbb{Z}_n$, with the property that summation and multiplication from $mathbb Z_n$ induces on $L$ a group structure.

I think that a proof ad-absurdum would provide a result, because the reciprocal seems easy to handle as I know that for a group $Z_k$, the invertible elements must be coprime with $k$ and I find so a ,,square-free" group. Any help to continue it?

abstract-algebra group-theory field-theory

share|cite|improve this question

edited Mar 15 at 21:34

user26857

39.4k124183

asked Mar 15 at 17:20

user651692

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The group structure induced on $L $ is additive or multiplicative?
  $endgroup$
  – Fabio Lucchini
  Mar 15 at 19:03
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Show that $ninmathbb{N}$ is square-free if and only if there is a subset different from the null set, $Lsubseteqmathbb{Z}_n$, with the property that summation and multiplication from $mathbb Z_n$ induces on $L$ a group structure.

I think that a proof ad-absurdum would provide a result, because the reciprocal seems easy to handle as I know that for a group $Z_k$, the invertible elements must be coprime with $k$ and I find so a ,,square-free" group. Any help to continue it?

abstract-algebra group-theory field-theory

share|cite|improve this question

edited Mar 15 at 21:34

user26857

39.4k124183

asked Mar 15 at 17:20

user651692

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The group structure induced on $L $ is additive or multiplicative?
  $endgroup$
  – Fabio Lucchini
  Mar 15 at 19:03
  

  
  
  
  

add a comment | 

$begingroup$

Show that $ninmathbb{N}$ is square-free if and only if there is a subset different from the null set, $Lsubseteqmathbb{Z}_n$, with the property that summation and multiplication from $mathbb Z_n$ induces on $L$ a group structure.

I think that a proof ad-absurdum would provide a result, because the reciprocal seems easy to handle as I know that for a group $Z_k$, the invertible elements must be coprime with $k$ and I find so a ,,square-free" group. Any help to continue it?

abstract-algebra group-theory field-theory

share|cite|improve this question

edited Mar 15 at 21:34

user26857

39.4k124183

asked Mar 15 at 17:20

user651692

$endgroup$

share|cite|improve this question

edited Mar 15 at 21:34

user26857

39.4k124183

asked Mar 15 at 17:20

user651692

share|cite|improve this question

edited Mar 15 at 21:34

user26857

39.4k124183

asked Mar 15 at 17:20

user651692

edited Mar 15 at 21:34

user26857

39.4k124183

edited Mar 15 at 21:34

user26857

39.4k124183

1 Answer
1

active

oldest

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2

$begingroup$

This question is completely incoherent. I have no idea what the intended question is supposed to be. Here are all the possible interpretations I can form of it, and why they don't work.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under addition.

Fails because $Bbb{Z}/(9)$ contains $(3)$.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under multiplication.

Fails because $Bbb{Z}/(9)$ contains $langle 4rangle$, which has order 3.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under addition and multiplication.

We can always take $L={0}$ to satisfy this, and if we require $Lne {0}$, then
this still makes no sense, because in order for $L$ to be a group under addition, it must contain $0$, but then if $xin L$, $xne 0$, we have $xcdot 0 = 0$, so $L$ cannot be a group under multiplication.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a field under addition and multiplication ($L$ is a group under addition, and $Lsetminus {0}$ is a group under multiplication).

Let $xin L$, $xne 0$. Then $x$ must be invertible in $Bbb{Z}/(n)$, which means that it is relatively prime to $Bbb{Z}/(n)$. Hence $x$ generates the additive group of $Bbb{Z}/(n)$, which implies that $L=Bbb{Z}/(n)$. Then $Bbb{Z}/(n)$ is a field if and only if $n$ is prime. Thus this doesn't work either.

share|cite|improve this answer

answered Mar 15 at 20:06

jgonjgon

16k32143

$endgroup$

add a comment | 

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2

$begingroup$

This question is completely incoherent. I have no idea what the intended question is supposed to be. Here are all the possible interpretations I can form of it, and why they don't work.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under addition.

Fails because $Bbb{Z}/(9)$ contains $(3)$.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under multiplication.

Fails because $Bbb{Z}/(9)$ contains $langle 4rangle$, which has order 3.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under addition and multiplication.

We can always take $L={0}$ to satisfy this, and if we require $Lne {0}$, then
this still makes no sense, because in order for $L$ to be a group under addition, it must contain $0$, but then if $xin L$, $xne 0$, we have $xcdot 0 = 0$, so $L$ cannot be a group under multiplication.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a field under addition and multiplication ($L$ is a group under addition, and $Lsetminus {0}$ is a group under multiplication).

Let $xin L$, $xne 0$. Then $x$ must be invertible in $Bbb{Z}/(n)$, which means that it is relatively prime to $Bbb{Z}/(n)$. Hence $x$ generates the additive group of $Bbb{Z}/(n)$, which implies that $L=Bbb{Z}/(n)$. Then $Bbb{Z}/(n)$ is a field if and only if $n$ is prime. Thus this doesn't work either.

share|cite|improve this answer

answered Mar 15 at 20:06

jgonjgon

16k32143

$endgroup$

add a comment | 

2

$begingroup$

This question is completely incoherent. I have no idea what the intended question is supposed to be. Here are all the possible interpretations I can form of it, and why they don't work.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under addition.

Fails because $Bbb{Z}/(9)$ contains $(3)$.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under multiplication.

Fails because $Bbb{Z}/(9)$ contains $langle 4rangle$, which has order 3.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under addition and multiplication.

We can always take $L={0}$ to satisfy this, and if we require $Lne {0}$, then
this still makes no sense, because in order for $L$ to be a group under addition, it must contain $0$, but then if $xin L$, $xne 0$, we have $xcdot 0 = 0$, so $L$ cannot be a group under multiplication.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a field under addition and multiplication ($L$ is a group under addition, and $Lsetminus {0}$ is a group under multiplication).

Let $xin L$, $xne 0$. Then $x$ must be invertible in $Bbb{Z}/(n)$, which means that it is relatively prime to $Bbb{Z}/(n)$. Hence $x$ generates the additive group of $Bbb{Z}/(n)$, which implies that $L=Bbb{Z}/(n)$. Then $Bbb{Z}/(n)$ is a field if and only if $n$ is prime. Thus this doesn't work either.

share|cite|improve this answer

answered Mar 15 at 20:06

jgonjgon

16k32143

$endgroup$

add a comment | 

$begingroup$

This question is completely incoherent. I have no idea what the intended question is supposed to be. Here are all the possible interpretations I can form of it, and why they don't work.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under addition.

Fails because $Bbb{Z}/(9)$ contains $(3)$.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under multiplication.

Fails because $Bbb{Z}/(9)$ contains $langle 4rangle$, which has order 3.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a group under addition and multiplication.

We can always take $L={0}$ to satisfy this, and if we require $Lne {0}$, then
this still makes no sense, because in order for $L$ to be a group under addition, it must contain $0$, but then if $xin L$, $xne 0$, we have $xcdot 0 = 0$, so $L$ cannot be a group under multiplication.

$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq Bbb{Z}/(n)$ such that $L$ is a field under addition and multiplication ($L$ is a group under addition, and $Lsetminus {0}$ is a group under multiplication).

Let $xin L$, $xne 0$. Then $x$ must be invertible in $Bbb{Z}/(n)$, which means that it is relatively prime to $Bbb{Z}/(n)$. Hence $x$ generates the additive group of $Bbb{Z}/(n)$, which implies that $L=Bbb{Z}/(n)$. Then $Bbb{Z}/(n)$ is a field if and only if $n$ is prime. Thus this doesn't work either.

share|cite|improve this answer

answered Mar 15 at 20:06

jgonjgon

16k32143

$endgroup$

share|cite|improve this answer

answered Mar 15 at 20:06

jgonjgon

16k32143

answered Mar 15 at 20:06

jgonjgon

16k32143

answered Mar 15 at 20:06

jgonjgon

16k32143