Let G be a finite abelian group. The exponent of G is defined exp(G) = max{ord(g

Question

Let G be a finite abelian group. The exponent of G is defined exp(G) = max{ord(g) | g is an element of H}, where ord(g) is the order of an element g in G. For this problem, standard results from group theory (such as Lagrange Theorem and the fact that element orders divide the group order) may be used. a) Prove that exp(G) divides the order |G| of G. ; b) Prove that exp(G) = lcm{ord(g) | g is an element of H} ; c) Give an example of a finite abelian group G for which exp(G) = |G|. Justify. ; d) Give an example of a finite abelian group G for which exp(G) does not equal |G|. Justify.

Explanation / Answer

I assume p is a prime number (the result is false if not: the cyclic group of order 6^2 = 36 has elements of orders that are not powers of 6, like 2 or 4 or 9).One direction is as you say: suppose G has order p^n for some nonnegative integer n, and let x be an element of G. By Lagrange's theorem the order of x must divide p^n. Since p is prime this means (e.g. using uniqueness of prime factorization of integers) that the order of x must also be a power of p. And x was arbitrary so the result fol

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