In proving G contains an element of order 15 if contains normal subgroups of orders 3 and 5, respectively, is $HK$ itself cyclic with order 15?

Question

There is an answer here, but it is a "roadmap". group containing normal subgroups of orders $3$ and $5$ contains element of order $15$ There are answers here, but they are "roadmaps" too. If $G$ contains normal subgroups of prime orders $p$ and $q$, then $G$ contains an element of order $pq$.

In the proofs given by Brian Bi linked here and by Alec Mouri linked here, it is concluded that $HK$ is a subgroup and is isomorphic to $H \times K$ and then that $HK$ contains an element of order 15.

1. Is it correct that it is either:

   • that the reason $HK$ contains an element of order 15 is that $HK$ is cyclic with order 15 or

   • that that's not reason $HK$ contains an element of order 15, but $HK$ is still cyclic with order 15?

2. Actually, are groups isomorphic to cyclic groups also cyclic with the same order?

I know if images of cyclic groups under surjective homomorphisms are cyclic, but I didn't think they would be of different orders if we also assume injective.

The following are screenshots of the proofs mentioned above (Kiefer Sutherland's voice)

Answer 1

As soon as you've proved that $HK\simeq H\times K$, you know that $H$ is the product of two groups of prime order, hence $H$ and $K$ are abelian (cyclic) groups, and as their orders are coprime, you can apply the Chinese remainder theorem, which asserts the product is cyclic with order the product of the orders of each factor.