All the $m$th powers commute with each other and all the $n$th powers commute with each other, $m$ and $n$ relative prime, is abelian.

Asked Mar 13 '19 at 15:27

Active Mar 13 '19 at 15:42

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Show that a group in which all the $m$th powers commute with each other and all the $n$th powers commute with each other ,$m$ and $n$ relatively prime, is abelian.

How to do this ? we know $1=mx+yn$ for $x,y \in \mathbb{Z}$

$ab=a^{mx}a^{ny}b^{yn}b^{mx}=a^{mx}b^{yn}a^{ny}b^{mx}$

but now how to switch $b^{yn}$ and $a^{mx}$ ??

edited Mar 13 '19 at 15:42

Bernard

175,478

asked Mar 13 '19 at 15:27

Eklavya

2,671

Found immediately in an https://approach0.xyz search. – rschwieb Mar 13 '19 at 15:51
take inverses to show those of powers $|m-n|$ commute. Then the proof parallels the proof of Bezout's lemma precisely. – fleablood Mar 13 '19 at 16:11

All the $m$th powers commute with each other and all the $n$th powers commute with each other, $m$ and $n$ relative prime, is abelian.

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