Let $C_k$ denote a cyclic group of order $k$. Show that if $m$ and $n$ are coprime, then $C_m \times C_n$ is cyclic and is isomorphic to $C_{mn}$. Show, however, that $C_3 \times C_3$ is not isomorphic to $C_9$.
So my questions are:
What is meant by the product of two groups $C_m \times C_n$ ?
How to solve this problem?
HINT: For the first part, show that the order of $C_m \times C_n$ is the same as the order of $C_{mn}$. Then use the fact that cyclic groups of the same order are isomorphic.
For the second part, what are the orders of elements in $C_3 \times C_3$? What about in $C_9$?
The direct product of groups $(G,\cdot)$ and $(H,\star)$, denoted by $G\times H$ is defined as follows: $$G\times H=\left\{(g,h)\,\vert\,g\in G,h\in H\right\}$$ with ($e_{\cdot}$ denotes the identity element): $$\begin{cases} (g,h)(g',h')&=(g\cdot g'\,,\,h\star h')\\ \text{Identity element} &: (e_{G},e_{H})\\ \text{Inverse of}\,\,(g,h) &: (g^{-1},h^{-1}) \end{cases}$$ As an information, there exists also a "semidirect product".
Let's take $g$ and $h$ the generators of $C_{m}$ and $C_{n}$, respectively. In your particular case, $C_{mn}=\left\{(g^{k},h^{l})\,\vert\,1\le k\le m,\,1\le l\le n\right\}$. You have to prove it is cyclic. Then, you take the morphism
$$\varphi:C_{m}\times C_{n}\to C_{mn}:(g,h)\mapsto gh$$ It can be easier to use that $C_{k}\cong \mathbb{Z}_{k}$ (and in $\mathbb{Z}_{k}$, $g^{a}=g+g+\cdots+g$ with $a$ elements in the sum).
For example, taking $\mathbb{Z}_{2}\times\mathbb{Z}_{3}$ and defining: $\varphi(a,b)=a+b$, you get $\varphi(0,3)=3$, $\varphi(3,2)=5$, $\varphi(3,4)=1$ and so on, until you get $\mathbb{Z}_{6}$.
Let $ \mathcal{C}_n=\langle a\rangle$ and $ \mathcal{C}_m=\langle b\rangle$. To show that $\mathcal{C}_n \times\mathcal{C}_m $ is cyclic, you should find a generator. Naturally, you would assume that generator is $(a,b)$. Now use Chinese remainder theorem to prove that indeed $\langle (a,b)\rangle = \mathcal{C}_n \times\mathcal{C}_m $.
Using the fact that $\mathcal{C}_m$ is isomorphic to $\mathbb{Z}_m$ and that order of $ \mathcal{C}_n \times\mathcal{C}_m $is $mn$ , it should not be hard to prove that $\mathcal{C}_n \times\mathcal{C}_m $ is isomorphic to $ \mathcal{C}_{mn}$.
And to show that $\mathcal{C}_3\times \mathcal{C}_3$ is not isomorphic to $\mathcal{C}_9$, it is enough to show that $\mathcal{C}_3\times \mathcal{C}_3$ is not cyclic. So verify that there is no element of order 9 in that group.
