Final answer:
In a finite Abelian group with order a power of a prime p, every element's order is also a power of p. This follows from Lagrange's Theorem, which states that the order of any subgroup (including cyclic subgroups generated by elements) divides the group's order. Since p is prime, any divisor must be a power of p.
Step-by-step explanation:
To demonstrate that if a finite Abelian group has order a power of a prime p, then the order of every element in the group is a power of p, we must use some properties of finite groups and the definition of the order of an element.
Let's consider a group G which is finite, Abelian, and its order (the number of elements in G) is a power of a prime p. That is, |G| = pn for some non-negative integer n.
Now, the order of an element a in G is the smallest positive integer m such that am = e, where e is the identity element of the group.
By Lagrange's Theorem in group theory, the order of any subgroup of G must divide the order of G. Since each element a in G generates a cyclic subgroup with the order equal to the order of the element, the order of that element must divide pn.
Therefore, the order of any element a must be a divisor of pn, which means the order of the element is also a power of p because p is a prime number and the divisors of a power of a prime are also powers of that prime.