MATH - GROUP

Please check your assumption:
Since x belongs to G which is a group, there exists an element, say y, in G such that xy=yx=e where e is the identity element of G. Assume ord(x) be n, (i.e. x^n=e and x^m ≠e for all m<n). Then we can show that if G is abelian, ord(y)=ord(x), which contradicts with your assumption that for all g belonging to G, ord(g) is less than ord(x).
(1) (y)^n= e(y)^n=[x^n][y^n] (since x^n=e)
=(xy)^n (since G is abelian)
=e^n (since xy=e)
=e
(2) Assume that there exists an integer m such that m<n and y^m=e,
then x^m= (x^m)e
=(x^m)(y^m) (substitute y^m=e)
= (xy)^m (since G is abelian)
=e^m
=e, which contradicts to the given information that n=ord(x).
By (1) and (2), ord(y)=n=ord(x).

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