Showing a group isn't cyclic Which of the following groups are cyclic? For each cyclic group, list all the generators of the group.
$$G_1 = \langle \mathbb{Z},+\rangle\;\;G_2 = \langle\mathbb{Q}, +\rangle\;\;G_3=\langle\mathbb{Q}^+, \cdot\rangle\;\;G_4 = \langle 6\mathbb{Z}, +\rangle$$
$$G_5 = \{6^n \mid n\in\mathbb{Z}\} \text{ under multiplication}$$
$$G_6 =\{a + b \sqrt{2}\mid a, b\in \mathbb{Z}\}\;\;\text{under addition} $$
My book says that $G_2$ and $G_3$ aren't cyclic, but it doesn't explain how they arrive to that conclusion. How exactly do you show that the groups aren't cyclic? In other words, how do I show that the group has no generator?
 A: HINT: For $G_2$, show that if $0\ne q\in\Bbb Q$, the group generated by $q$ does not contain $q/2$. 
For $G_3$ you can use the same idea: show that if $0<q\in\Bbb Q$, there is some $r\in\Bbb Q^+$ that is not in $\langle q\rangle$, the group generated by $q$. First work out just what is in $\langle q\rangle$; once you’ve done that, it’s not too hard to come up with something in $\Bbb Q^+\setminus\langle q\rangle$.
In both cases what you’re doing is showing that no element of the group generates the whole group, which therefore cannot be cyclic. It may help to notice that in both $G_2$ and $G_3$ the group generated by $q$ has lots of gaps.
A: Generally speaking: suppose there was a generator and derive a contradiction.
(Alternative methods include, for example, showing the group is isomorphic to another group which you already know to be non-cyclic.)
For $G_2$, suppose $a$ is the generator. Will you ever be able to get the rational number $3a/2$? (That is, $a^1 = a$ and $a^2 = a + a = 2a$; but what about the point halfway between $a$ and $2a$?)
For $G_3$, suppose without loss of generality that $a > 1$ is the generator. (If $b < 1$ is a generator, then so is $b' = 1/b > 1$.) Then we can generate $a$ and $a^2$, but what about the point halfway between them? That is, what about $(a + a^2)/2$?
A: We start off by defining an action of $\Bbb{Z}$ on $G_2$. Fix a natural number $a$ which is neither $0$ nor $1$. For $n\in\mathbb{Z}$, $x\in\mathbb{Q}$ let $n\cdot x=a^nx$, where the juxtaposition notation $mx$ stands for $\sum_{i=1}^mx$. I claim this is an action via automorphisms $\varphi:\mathbb{Z}\to\text{Aut}G_2$.
To verify this, we first show that it is an action. If $m,n$ are integers and $x$ is rational, then $$n\cdot (m\cdot x)=n\cdot(a^mx)=(a^na^mx)=(n+m)\cdot x.$$
Next, we need to show that each map "$m\cdot$" is a homomorphism, let $y$ be any rational number:$$n\cdot(x+y)=a^n(x+y)=a^nx+a^ny=n\cdot x+n\cdot y.$$ Having verified that $\varphi$ is a homomorphism we use it show that $G_2$ cannot be cyclic.
Obviously, the kernel is trivial, so the image of $\varphi$ infinite. If $G_2$ were cyclic, though, the group of automorphisms on $G_2$ would consist of just two elements namely identity and inversion. Therefore, $G_2$ is not cyclic.
Note that a similar argument does not prove $G_3$ is not cyclic.
