Let G be group, a ɛ G and O(a)=n. Then for any positive integer k, O(a^k) =n/(n,k).

G is group and a,b be elements of G such that (o(a),o(b))=1 and ab=ba then o(ab)=o(a)o(b).

Every subgroup of a cyclic group is cyclic.

Let G be a group, a ɛ G and O(a)=n. Then a^m=e iff n|m.

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If a and b are two elements of a.finite group G s.t.ab=ba and (o(a),o(b))=1 then o(ab)=o(a)o(b)

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#04 trick of abstarct algebra | If in a group G , a^5=e ,aba^-1 =b^2 , then order of b is

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[a^k] = [a^gcd(n,k)] and |a^k| = n/gcd(n,k) Proof (Abstract Algebra)

O(ab)=O(a)*O(b) where g.c.d(O(a),O(b))=1 and ab=ba.

Find the order of each element of the group {0,1,2,3,4,5} under addition modulo 6