Is $(ab)^n=a^nb^n$ condition on finite group implies that group to be abelian?

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I was encountered following problem :
Let G be the finite abelian group with order not divisible by 3 also $forall a,bin G$ such that $(ab)^3=a^3b^3$ then G is abelian.

This I am able to prove using the following

$(ab)^3$=$a^3b^3$ which implies $(ba)^2=a^2b^2$
Now $(ab)^4=((ab)^2)^2$=$((b^2a^2))^2=(a^4b^4)$

$(ab)^4=abababab=a^4b^4$
That implies $(ba)^3=a^3b^3=(ab)^3$

There is automorphism defined as $phi:Gto G$ as$phi(g)=g^3$ as order of group is not divisible by 3.




I am intersted in knowing can we generalise this to any n ?If not then what additional assumption are required us to make it work?



i.e. G is the finite group with order not divisible by n$(ab)^n=a^nb^n$ then G is abelian group




I had tried but I get recursive step from which I could not conclude anything. Any help will be appreciated




abstract-algebra group-theory finite-groups




share|cite|improve this question















  • 2




    You need a stronger condition, otherwise suppose the order of the group is $k$ and let $n=2k$.
    – lulu
    Aug 1 at 13:37






  • 1




    math.stackexchange.com/questions/928772/… seems to be relevant
    – Nicky Hekster
    Aug 1 at 13:46










  • See also this question.
    – Dietrich Burde
    Aug 1 at 14:15











  • See the notion of $n$-abelian groups.
    – Arnaud Mortier
    Aug 1 at 14:47














up vote
0
down vote

favorite












I was encountered following problem :
Let G be the finite abelian group with order not divisible by 3 also $forall a,bin G$ such that $(ab)^3=a^3b^3$ then G is abelian.

This I am able to prove using the following

$(ab)^3$=$a^3b^3$ which implies $(ba)^2=a^2b^2$
Now $(ab)^4=((ab)^2)^2$=$((b^2a^2))^2=(a^4b^4)$

$(ab)^4=abababab=a^4b^4$
That implies $(ba)^3=a^3b^3=(ab)^3$

There is automorphism defined as $phi:Gto G$ as$phi(g)=g^3$ as order of group is not divisible by 3.




I am intersted in knowing can we generalise this to any n ?If not then what additional assumption are required us to make it work?



i.e. G is the finite group with order not divisible by n$(ab)^n=a^nb^n$ then G is abelian group




I had tried but I get recursive step from which I could not conclude anything. Any help will be appreciated




abstract-algebra group-theory finite-groups




share|cite|improve this question















  • 2




    You need a stronger condition, otherwise suppose the order of the group is $k$ and let $n=2k$.
    – lulu
    Aug 1 at 13:37






  • 1




    math.stackexchange.com/questions/928772/… seems to be relevant
    – Nicky Hekster
    Aug 1 at 13:46










  • See also this question.
    – Dietrich Burde
    Aug 1 at 14:15











  • See the notion of $n$-abelian groups.
    – Arnaud Mortier
    Aug 1 at 14:47












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I was encountered following problem :
Let G be the finite abelian group with order not divisible by 3 also $forall a,bin G$ such that $(ab)^3=a^3b^3$ then G is abelian.

This I am able to prove using the following

$(ab)^3$=$a^3b^3$ which implies $(ba)^2=a^2b^2$
Now $(ab)^4=((ab)^2)^2$=$((b^2a^2))^2=(a^4b^4)$

$(ab)^4=abababab=a^4b^4$
That implies $(ba)^3=a^3b^3=(ab)^3$

There is automorphism defined as $phi:Gto G$ as$phi(g)=g^3$ as order of group is not divisible by 3.




I am intersted in knowing can we generalise this to any n ?If not then what additional assumption are required us to make it work?



i.e. G is the finite group with order not divisible by n$(ab)^n=a^nb^n$ then G is abelian group




I had tried but I get recursive step from which I could not conclude anything. Any help will be appreciated




abstract-algebra group-theory finite-groups




share|cite|improve this question











I was encountered following problem :
Let G be the finite abelian group with order not divisible by 3 also $forall a,bin G$ such that $(ab)^3=a^3b^3$ then G is abelian.

This I am able to prove using the following

$(ab)^3$=$a^3b^3$ which implies $(ba)^2=a^2b^2$
Now $(ab)^4=((ab)^2)^2$=$((b^2a^2))^2=(a^4b^4)$

$(ab)^4=abababab=a^4b^4$
That implies $(ba)^3=a^3b^3=(ab)^3$

There is automorphism defined as $phi:Gto G$ as$phi(g)=g^3$ as order of group is not divisible by 3.




I am intersted in knowing can we generalise this to any n ?If not then what additional assumption are required us to make it work?



i.e. G is the finite group with order not divisible by n$(ab)^n=a^nb^n$ then G is abelian group




I had tried but I get recursive step from which I could not conclude anything. Any help will be appreciated





abstract-algebra group-theory finite-groups





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 1 at 13:34









SRJ

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  • 2




    You need a stronger condition, otherwise suppose the order of the group is $k$ and let $n=2k$.
    – lulu
    Aug 1 at 13:37






  • 1




    math.stackexchange.com/questions/928772/… seems to be relevant
    – Nicky Hekster
    Aug 1 at 13:46










  • See also this question.
    – Dietrich Burde
    Aug 1 at 14:15











  • See the notion of $n$-abelian groups.
    – Arnaud Mortier
    Aug 1 at 14:47












  • 2




    You need a stronger condition, otherwise suppose the order of the group is $k$ and let $n=2k$.
    – lulu
    Aug 1 at 13:37






  • 1




    math.stackexchange.com/questions/928772/… seems to be relevant
    – Nicky Hekster
    Aug 1 at 13:46










  • See also this question.
    – Dietrich Burde
    Aug 1 at 14:15











  • See the notion of $n$-abelian groups.
    – Arnaud Mortier
    Aug 1 at 14:47







2




2




You need a stronger condition, otherwise suppose the order of the group is $k$ and let $n=2k$.
– lulu
Aug 1 at 13:37




You need a stronger condition, otherwise suppose the order of the group is $k$ and let $n=2k$.
– lulu
Aug 1 at 13:37




1




1




math.stackexchange.com/questions/928772/… seems to be relevant
– Nicky Hekster
Aug 1 at 13:46




math.stackexchange.com/questions/928772/… seems to be relevant
– Nicky Hekster
Aug 1 at 13:46












See also this question.
– Dietrich Burde
Aug 1 at 14:15





See also this question.
– Dietrich Burde
Aug 1 at 14:15













See the notion of $n$-abelian groups.
– Arnaud Mortier
Aug 1 at 14:47




See the notion of $n$-abelian groups.
– Arnaud Mortier
Aug 1 at 14:47















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