Question

A group has exactly m distinct subgroups of prime order p. Prove that the total no. Of elements of order p is m(p-1)

Answer 1

My solution is almost correct, however there is a fault in my reasoning:

A cyclic group of order n doesn't have just 1 generator, it has ϕ(n) generators where ϕ(n) is Euler's phi function. Or in other words: The amount of numbers which are coprime to the order of the cyclic group is equal to the amount of generators for that cyclic group.

Since p is prime we have p−1 such numbers, thus we have m(p−1) such elements in G.