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Show that $n$ is a prime number

Square free finite abelian group is cyclic$G$ is an abelian group of order a product of distinct primes $implies G$ is cyclic?Finite abelian groups and subgroups.Show that $G$ is cyclicProof for prime avoidance lemma.Showing abelian group order that is the product of primes is cyclicProof that multiplication of two numbers plus $2$ is a prime numberProve that there is an infinite number of prime numbersIf $|G| = p_1^{r_1}…p_k^{r_k}$ it is the direct internal product of its Sylow subgroupsOrder of an arbitrary (finite) product of subgroups $|N_1cdots N_k|$

1

1

$begingroup$

Let $ninmathbb{N}$, $ngeq2$ and $(G,.)$ an abelian group of order $n$, with the property that for any $ain G$, the number of endomorphisms of $G$ with $f(a)=a$ is equal to $n+1-ord(a)$, where $ord(a)$ is the order of $a$. Show that $n$ is a prime number.

My approach is to suppose that $n$ is not prime, so there is a decomposition for $n=p_1p_2...p_k$, where $p_1,dots ,p_k$ are prime numbers and from Cauchy's theorem there is an $ain G$ such that $ord(a)=p_i$. Now $f(e)=f(a^{p_i})=(f(a))^{p_i}=e$, and so the number of endomorphisms $f$ is $n+1-1=n$.

However, from the hypothesis, the number of $f$ is $n+1-p_i$, and so $n+1-p_i=n$ $Rightarrow p_i=1$. As we chosen $p_i$ arbitrarily, we get that all the numbers $p_i,iin{0,1,...,k}$ are $1Rightarrow n=1$, false with $ngeq2$.

Is my proof correct?

abstract-algebra group-theory proof-verification finite-groups

share|cite|improve this question

edited Mar 9 at 18:46

Shaun

9,366113684

asked Mar 9 at 17:57

Jacob DeniculaJacob Denicula

554

New contributor

Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The problem seems ill-formed- wouldn't $f(e)=e$ always hold for any endomorphism fixing an element (regardless of the primality of $|G|$)?
  $endgroup$
  – Cardioid_Ass_22
  Mar 9 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Cardioid_Ass_22 but that needs to hold for all elements.
  $endgroup$
  – enedil
  Mar 9 at 19:49
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Let $ninmathbb{N}$, $ngeq2$ and $(G,.)$ an abelian group of order $n$, with the property that for any $ain G$, the number of endomorphisms of $G$ with $f(a)=a$ is equal to $n+1-ord(a)$, where $ord(a)$ is the order of $a$. Show that $n$ is a prime number.

My approach is to suppose that $n$ is not prime, so there is a decomposition for $n=p_1p_2...p_k$, where $p_1,dots ,p_k$ are prime numbers and from Cauchy's theorem there is an $ain G$ such that $ord(a)=p_i$. Now $f(e)=f(a^{p_i})=(f(a))^{p_i}=e$, and so the number of endomorphisms $f$ is $n+1-1=n$.

However, from the hypothesis, the number of $f$ is $n+1-p_i$, and so $n+1-p_i=n$ $Rightarrow p_i=1$. As we chosen $p_i$ arbitrarily, we get that all the numbers $p_i,iin{0,1,...,k}$ are $1Rightarrow n=1$, false with $ngeq2$.

Is my proof correct?

abstract-algebra group-theory proof-verification finite-groups

share|cite|improve this question

edited Mar 9 at 18:46

Shaun

9,366113684

asked Mar 9 at 17:57

Jacob DeniculaJacob Denicula

554

New contributor

Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The problem seems ill-formed- wouldn't $f(e)=e$ always hold for any endomorphism fixing an element (regardless of the primality of $|G|$)?
  $endgroup$
  – Cardioid_Ass_22
  Mar 9 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Cardioid_Ass_22 but that needs to hold for all elements.
  $endgroup$
  – enedil
  Mar 9 at 19:49
  

  
  
  
  

add a comment | 

$begingroup$

Let $ninmathbb{N}$, $ngeq2$ and $(G,.)$ an abelian group of order $n$, with the property that for any $ain G$, the number of endomorphisms of $G$ with $f(a)=a$ is equal to $n+1-ord(a)$, where $ord(a)$ is the order of $a$. Show that $n$ is a prime number.

My approach is to suppose that $n$ is not prime, so there is a decomposition for $n=p_1p_2...p_k$, where $p_1,dots ,p_k$ are prime numbers and from Cauchy's theorem there is an $ain G$ such that $ord(a)=p_i$. Now $f(e)=f(a^{p_i})=(f(a))^{p_i}=e$, and so the number of endomorphisms $f$ is $n+1-1=n$.

However, from the hypothesis, the number of $f$ is $n+1-p_i$, and so $n+1-p_i=n$ $Rightarrow p_i=1$. As we chosen $p_i$ arbitrarily, we get that all the numbers $p_i,iin{0,1,...,k}$ are $1Rightarrow n=1$, false with $ngeq2$.

Is my proof correct?

abstract-algebra group-theory proof-verification finite-groups

share|cite|improve this question

edited Mar 9 at 18:46

Shaun

9,366113684

asked Mar 9 at 17:57

Jacob DeniculaJacob Denicula

554

New contributor

Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 9 at 18:46

Shaun

9,366113684

asked Mar 9 at 17:57

Jacob DeniculaJacob Denicula

554

New contributor

Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 9 at 18:46

Shaun

9,366113684

asked Mar 9 at 17:57

Jacob DeniculaJacob Denicula

554

New contributor

Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 9 at 18:46

Shaun

9,366113684

edited Mar 9 at 18:46

Shaun

9,366113684

asked Mar 9 at 17:57

Jacob DeniculaJacob Denicula

554

New contributor

Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 9 at 17:57

Jacob DeniculaJacob Denicula

554

1 Answer
1

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-1

$begingroup$

You haven't used the hypothesis that the group is abelian.

I corrected some points of grammar & punctuation, though, too, like not starting a sentence with a mathematical symbol, adding a space after each comma, etc.

share|cite|improve this answer

edited Mar 9 at 19:07

answered Mar 9 at 18:54

ShaunShaun

9,366113684

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! How should I use the fact that the group is abelian?
  $endgroup$
  – Jacob Denicula
  Mar 9 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Based on what I understand, you don't need that hypothesis. This is a bad sign though. I'm inclined to think I'm wrong. Where is the problem from, @JacobDenicula?
  $endgroup$
  – Shaun
  Mar 9 at 19:27
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A mathematical magazine. It doesn`t matter as it has no proof.
  $endgroup$
  – Jacob Denicula
  Mar 9 at 19:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, it doesn't, epistemologically, @JacobDenicula. (Try to use ' instead of ` on this site for technical reasons.) If it has no proof, it has no proof. But to see whether it's likely to be correct, it helps to know from whence it came.
  $endgroup$
  – Shaun
  Mar 9 at 19:59
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
  $endgroup$
  – Alan Muniz
  Mar 9 at 21:57
  

  
  
  
  

 | 
show 1 more comment

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-1

$begingroup$

You haven't used the hypothesis that the group is abelian.

I corrected some points of grammar & punctuation, though, too, like not starting a sentence with a mathematical symbol, adding a space after each comma, etc.

share|cite|improve this answer

edited Mar 9 at 19:07

answered Mar 9 at 18:54

ShaunShaun

9,366113684

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! How should I use the fact that the group is abelian?
  $endgroup$
  – Jacob Denicula
  Mar 9 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Based on what I understand, you don't need that hypothesis. This is a bad sign though. I'm inclined to think I'm wrong. Where is the problem from, @JacobDenicula?
  $endgroup$
  – Shaun
  Mar 9 at 19:27
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A mathematical magazine. It doesn`t matter as it has no proof.
  $endgroup$
  – Jacob Denicula
  Mar 9 at 19:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, it doesn't, epistemologically, @JacobDenicula. (Try to use ' instead of ` on this site for technical reasons.) If it has no proof, it has no proof. But to see whether it's likely to be correct, it helps to know from whence it came.
  $endgroup$
  – Shaun
  Mar 9 at 19:59
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
  $endgroup$
  – Alan Muniz
  Mar 9 at 21:57
  

  
  
  
  

 | 
show 1 more comment

-1

$begingroup$

You haven't used the hypothesis that the group is abelian.

I corrected some points of grammar & punctuation, though, too, like not starting a sentence with a mathematical symbol, adding a space after each comma, etc.

share|cite|improve this answer

edited Mar 9 at 19:07

answered Mar 9 at 18:54

ShaunShaun

9,366113684

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! How should I use the fact that the group is abelian?
  $endgroup$
  – Jacob Denicula
  Mar 9 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Based on what I understand, you don't need that hypothesis. This is a bad sign though. I'm inclined to think I'm wrong. Where is the problem from, @JacobDenicula?
  $endgroup$
  – Shaun
  Mar 9 at 19:27
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A mathematical magazine. It doesn`t matter as it has no proof.
  $endgroup$
  – Jacob Denicula
  Mar 9 at 19:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, it doesn't, epistemologically, @JacobDenicula. (Try to use ' instead of ` on this site for technical reasons.) If it has no proof, it has no proof. But to see whether it's likely to be correct, it helps to know from whence it came.
  $endgroup$
  – Shaun
  Mar 9 at 19:59
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review
  $endgroup$
  – Alan Muniz
  Mar 9 at 21:57
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

You haven't used the hypothesis that the group is abelian.

I corrected some points of grammar & punctuation, though, too, like not starting a sentence with a mathematical symbol, adding a space after each comma, etc.

share|cite|improve this answer

edited Mar 9 at 19:07

answered Mar 9 at 18:54

ShaunShaun

9,366113684

$endgroup$

share|cite|improve this answer

edited Mar 9 at 19:07

answered Mar 9 at 18:54

ShaunShaun

9,366113684

answered Mar 9 at 18:54

ShaunShaun

9,366113684

answered Mar 9 at 18:54

ShaunShaun

9,366113684