Hint: an abelian group of order with primes, is cyclic.Hello, I am stuck with the following question and was wondering if anyone could give me some advice.
Let be an abelian group of order pq, where p<q are primes. Show that there are exactly p-1 elements of order q, q-1 elements of order q, and (p-1)(q-1) elements of order pq.
OK so we have a theorem which tells us that there must be only one subgroup of order q (which is normal). This subgroup is cyclic because it is of prime order, so every non identity element is of order q. This gives q-1 elements of order q. There can be no other elements of order q in G since they would generate a distinct subgroup of order q, and there can be only one.
So there are definitely only q-1 elements of order q in G.
By Cauchy's theorem, there exists an element of order p in G. This generates a cyclic group of order p which has p-1 elements of order p. why can there be no more than that?
Knowing this part implies the number of elements of order pq = pq - (p-1) - (q-1) = (p-1)(q-1), so how do I show that there are no more than p-1 elements of order p?
Any help with this would be appreciated, Thank you