How to properly prove a group is cyclic? How does one properly prove a group is cyclic? For example the group $\mathbb{Z}$. And how does one prove something is not cyclic, like $(\mathbb{Q}^{+},.)$?
 A: A group $G$ is cyclic when $G = \langle a\rangle = \{a^n : n\in \mathbb{Z}\}$ (written multiplicatively) for some $a\in G$.
Written additively, we have $\langle a \rangle = \{an: n\in\mathbb{Z}\}$.
So to show that $\mathbb{Z}$ is cyclic you just note that
$$
\mathbb{Z} = \{1\cdot n : n\in \mathbb{Z}\}.
$$
To show that $\mathbb{Q}$ is not a cyclic group you could assume that it is cyclic and then derive a contradiction.
So say that $\frac{a}{b}$ (reduced fraction) is a generator for $\mathbb{Q}$. That is
$$
\mathbb{Q} = \langle \frac{a}{b} \rangle = \{\frac{a}{b}n: n\in \mathbb{Z}\}.
$$
Now $\frac{1}{b}\in \mathbb{Q}$ so $\frac{an}{b} = \frac{1}{b}$ for some $n$. This forces $a = 1$. Now just note that $\frac{1}{2b}\not\in\langle \frac{1}{b}\rangle$.
EDIT: I see now that I mistook your $\mathbb{Q}^+$ for the set group of rational numbers under addition. I will update my answer.
So let's consider the group of positive rational numbers $\mathbb{Q}^+$ under multiplication. Assume again that $\mathbb{Q}$ is cyclic. Then, again, there is a $\frac{a}{b}$ (reduced) such that
$$
\mathbb{Q} = \langle \frac{a}{b} \rangle = \{\left(\frac{a}{b}\right)^n : n\in \mathbb{Z}\}.
$$
Now $\frac{1}{b}\in \mathbb{Q}$ so there is an $n\in \mathbb{Z}$ such that
$$
\frac{1}{b} = \frac{a^n}{b^n}\;\;\;\; \text{(still reduced)}.
$$
So
$$
b^{n-1} = a^{n}.
$$
This implies that $a = b = 1$ since $\frac{a}{b}$ is reduced (i.e. $\gcd(a,b) = 1$.
(NOTE: This is just one way to do this. It isn't necessarily pretty, but you can probably now find a prettier way to do it.)
