This exercise generalizes the previous two. Let (G,∗) and (H,⋆) be groups. Their direct product is a new group (G,∗)×(H,⋆) whose elements are all ordered pairs (g,h) where g∈G and h∈H. The operation for this group (let's use the symbol ⋅) is defined by
(g1,h1)⋅(g2,h2)=(g1∗g2,h1⋆h2)
For example, let's consider (Z5∗,⊗)×(Z3,⊕). We have
Z5∗={1,2,3,4} and Z3={0,1,2}
and therefore the elements of (Z5∗,⊗)×(Z3,⊕) are
{(1,0),(1,1),(1,2),(2,0),(2,1),(2,2),(3,0),(3,1),(3,2),(4,0),(4,1),(4,2)}
The operation (g1,h1)⋅(g2,h2) produces the value (g,h) where g=g1⊗g2 (operating in Z5∗) and h=h1⊕h2 (operating in Z3). For example,
(2,1)⋅(3,2)=(2⊗3,1⊕2)=(1,0)
You should convince yourself that if (G,∗) and (H,⋆) are groups, then (G,∗)×(H,⋆) is also a group. [Optional: Write a formal proof of this.]
We now come to the point of this problem: If (G,∗) and (H,⋆) are finite cyclic groups, then sometimes (G,∗)×(H,⋆) is cyclic and sometimes it is not. The question is, under what condition(s) is the direct product of two finite cyclic groups also cyclic?