If the element 'a' of a group G is order n, then a^m= e iff n is a divisor of m

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Let G be a group, a ɛ G and O(a)=n. Then a^m=e iff n|m. Let G be a group, a ɛ G and O(a)=n. Then a^m=e iff n|m.
prove that a^m=e iff n is divisor of m prove that a^m=e iff n is divisor of m
theorem group theory :-If order of element a of group G is n then, a^m=e, iff m is multiple of n theorem group theory :-If order of element a of group G is n then, a^m=e, iff m is multiple of n
If the element If the element 'a' of a group G is of order n, then a^m =e iff n is a divisor of m || Group Theory.
a^(m) is a generator of group G of order n iff GCD(m,n)=1 a^(m) is a generator of group G of order n iff GCD(m,n)=1
GROUP THEORY BSc L#52/If the element a of group G is of order n then am = e if and only if n is divi GROUP THEORY BSc L#52/If the element a of group G is of order n then am = e if and only if n is divi
If G is a group &a€G has order n,then a^m=1 iff n|m||group theory ||theorems||important for exams If G is a group &a€G has order n,then a^m=1 iff n|m||group theory ||theorems||important for exams
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