In understanding if there exist an abelian group of order $15$ that is not cyclic, I came up with this example: the direct product $\mathbb{Z}/3 \times \mathbb{Z}/5$. I know that this group is abelian, but I'm unsure if its cyclic or not.
If I can find one element in this set where I multiply it by a number that is not $15$, and I get $(\overline{0},\overline{0})$ then it can't be cyclic and I'm done, I think this is the case at least, is my logic correct, or would I need some number less than 15, multiply every element in this set by a number strictly less than 15, and get $(\overline{0},\overline{0})$. If this is the case then no such example exists.
Best Answer
Hint: What is the order of the element $(1,1) \in \mathbb{Z}_3 \times \mathbb{Z}_5?$ (there is a general formula for the order of an element in a direct product, but you can check it what the order is here by hand if you don't know this formula). What can you conclude about the group?
It can be shown that any group of order $15$ is isomorphic to $\mathbb{Z}_{15}$. This is usually proven by Sylow theory.