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    Order of the product of two elements in a group

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    Let G be a group (finite or infinite) and let a and b in G. Let o(a) represent the order of a.

    Suppose G is abelian and both a and b are of finite order. Show that ab is of finite order and o(ab) divides o(a)o(b).

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    Solution Preview

    The key point here is that G is abelian.
    Let n = o(a), m = o(b)
    Then a^n=e, b^m = e
    Now, (ab)^{nm} = (here you use the fact that G is abelian.) = a^{nm}b^{nm} = (a^n)^m ...

    Solution Summary

    A rigorous proof of the statement is given.