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## Homework Statement

EDIT: Necessary and sufficient condition for a^m = e -> a=e for a in a finite abelian group of order n, and I think the answer is (m,n)=1, but it is hard for me to show a^m =e -> a=e requires (m,n) =1.

## Homework Equations

## The Attempt at a Solution

I know that if GCD(m,n) =1 then if a^m = e, then o(a) | m, but by Lagrange, o(a) | n -> 1 = GCD(m,n) >= o(a) -> o(a) =1 -> a=e.

However, if a^m=e -> a=e, I thought the following: suppose d=GCD(m,n). Then, d= bm + cn (b,c integers) suppose a^m = e -> a^bm = e -> a^(bm + cn) = e -> a^d =e, then I thought so now we need a^d =e -> a=e, so o(a) can't divide d, but o(a) divides m and n, since a^m =e, whence o(a) | bm +cn -> o(a)|d

(but if o(a)|m, then a^m =e doesn't imply a= e) so I don't know if it is right to say o(a)|m, so I am stuck here and any help would be appreciated.

Thanks

EDIT: Necessary and sufficient condition for a^m = e -> a=e for a in a finite abelian group of order n, and I think the answer is (m,n)=1, but it is hard for me to show a^m =e -> a=e requires (m,n) =1.

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