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True/false: If two finite abelian groups of the same order have equal sum of element orders, then they are isomorphic?

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5

4

$begingroup$

Is it true that: If two finite abelian groups of the same order have equal sum of element orders, then they are isomorphic?

I have tried getting a counterexample, but failed.

So as not to be misunderstood by the term "sum of element orders," I will explain with an example.

The sum of element orders of $mathbbZ_2^2$ is $1+2+2+2=7$ while that of $mathbbZ_4$ is $1+2+4+4=11$. This example shows that the sum of element orders of $mathbbZ_2^2$ and $mathbbZ_4$ are not equal.

Looking forward to suggestions on how to prove the above highlighted proposition.

Thanks.

group-theory finite-groups group-isomorphism

share|cite|improve this question

edited yesterday

Shaun

9,334113684

asked Nov 21 '16 at 13:06

ChrisDavChrisDav

264

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's a fundamental theorem for finite abelian groups, which is like factorization in some sense. If your group is order $n=p_1^alpha_1p_2^alpha_2cdots p_k^alpha_k$, $p$'s distinct primes, then the finite abelian groups of order $n$ are $Z_q_1times Z_q_2timescdotstimes Z_q_l $ where each of the $q$'s are prime powers and the product of the $q$'s is $n$. This would at least let you(others) check specific $n$ for counterexamples if they exist.
  $endgroup$
  – snulty
  Nov 21 '16 at 13:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This appears to be a place to start: math.uaic.ro/~martar/pdf/articole/articol%2056.pdf
  $endgroup$
  – Prahlad Vaidyanathan
  Nov 21 '16 at 13:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ snuity, I know the result. In fact, I am sure we can't get a counter example up to abelian groups of order 5000 which I have carefully checked. So I believe the proposition is true.
  $endgroup$
  – ChrisDav
  Nov 21 '16 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PrahladVaidyanathan judging by the conjecture $6$ (op's question) and that it was computer checked for groups of order less than 100,000, I guess 'small' counterexamples won't be found.
  $endgroup$
  – snulty
  Nov 21 '16 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChrisDav my last comment also.
  $endgroup$
  – snulty
  Nov 21 '16 at 13:34
  

  
  
  
  

 | 
show 2 more comments

5

4

$begingroup$

Is it true that: If two finite abelian groups of the same order have equal sum of element orders, then they are isomorphic?

I have tried getting a counterexample, but failed.

So as not to be misunderstood by the term "sum of element orders," I will explain with an example.

The sum of element orders of $mathbbZ_2^2$ is $1+2+2+2=7$ while that of $mathbbZ_4$ is $1+2+4+4=11$. This example shows that the sum of element orders of $mathbbZ_2^2$ and $mathbbZ_4$ are not equal.

Looking forward to suggestions on how to prove the above highlighted proposition.

Thanks.

group-theory finite-groups group-isomorphism

share|cite|improve this question

edited yesterday

Shaun

9,334113684

asked Nov 21 '16 at 13:06

ChrisDavChrisDav

264

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's a fundamental theorem for finite abelian groups, which is like factorization in some sense. If your group is order $n=p_1^alpha_1p_2^alpha_2cdots p_k^alpha_k$, $p$'s distinct primes, then the finite abelian groups of order $n$ are $Z_q_1times Z_q_2timescdotstimes Z_q_l $ where each of the $q$'s are prime powers and the product of the $q$'s is $n$. This would at least let you(others) check specific $n$ for counterexamples if they exist.
  $endgroup$
  – snulty
  Nov 21 '16 at 13:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This appears to be a place to start: math.uaic.ro/~martar/pdf/articole/articol%2056.pdf
  $endgroup$
  – Prahlad Vaidyanathan
  Nov 21 '16 at 13:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ snuity, I know the result. In fact, I am sure we can't get a counter example up to abelian groups of order 5000 which I have carefully checked. So I believe the proposition is true.
  $endgroup$
  – ChrisDav
  Nov 21 '16 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PrahladVaidyanathan judging by the conjecture $6$ (op's question) and that it was computer checked for groups of order less than 100,000, I guess 'small' counterexamples won't be found.
  $endgroup$
  – snulty
  Nov 21 '16 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChrisDav my last comment also.
  $endgroup$
  – snulty
  Nov 21 '16 at 13:34
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Is it true that: If two finite abelian groups of the same order have equal sum of element orders, then they are isomorphic?

I have tried getting a counterexample, but failed.

So as not to be misunderstood by the term "sum of element orders," I will explain with an example.

The sum of element orders of $mathbbZ_2^2$ is $1+2+2+2=7$ while that of $mathbbZ_4$ is $1+2+4+4=11$. This example shows that the sum of element orders of $mathbbZ_2^2$ and $mathbbZ_4$ are not equal.

Looking forward to suggestions on how to prove the above highlighted proposition.

Thanks.

group-theory finite-groups group-isomorphism

share|cite|improve this question

edited yesterday

Shaun

9,334113684

asked Nov 21 '16 at 13:06

ChrisDavChrisDav

264

$endgroup$

share|cite|improve this question

edited yesterday

Shaun

9,334113684

asked Nov 21 '16 at 13:06

ChrisDavChrisDav

264

share|cite|improve this question

edited yesterday

Shaun

9,334113684

asked Nov 21 '16 at 13:06

ChrisDavChrisDav

264

edited yesterday

Shaun

9,334113684

edited yesterday

Shaun

9,334113684

asked Nov 21 '16 at 13:06

ChrisDavChrisDav

264

asked Nov 21 '16 at 13:06

ChrisDavChrisDav

264