Discussion
Consider the following statements :
(I) Let H and K be subgroups of an abelian group G. Suppose \(|H| = m\) and \(|K| = n\). Let \(d = \text{lcm}(m, n)\). Then G has a subgroup of order \(d\).
(II) The group of order 15 is abelian.
Which of the following is true ?
Answer: C
Both (I) and (II) are true.
(I) In an abelian group, HK is a subgroup, and it is finite of order \(mn/|H \cap K|\), a number divisible by both \(m\) and \(n\), hence by \(d = \text{lcm}(m, n)\). A finite abelian group has a subgroup of every order dividing its order (the converse of Lagrange holds for abelian groups), so HK, and therefore G, has a subgroup of order \(d\).
(II) Every group of order 15 is cyclic, hence abelian. By Sylow's theorems the number of Sylow 5-subgroups divides 3 and is 1 mod 5, so it is 1; the number of Sylow 3-subgroups divides 5 and is 1 mod 3, so it is 1. Both are normal and intersect trivially, so \(G \cong \mathbf{Z}_{3} \times \mathbf{Z}_{5} \cong \mathbf{Z}_{15}\).
So the options that deny either statement are wrong.
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