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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?

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    @amWhy: I’ve also lear$n$ed almost all of what I know about it here; I finally know enough to have at least some idea of what to go looking for if I *don’t* know how to do something.2012-11-05

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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.

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    @nonpop It was wishful thinking that we could get the additive group of integers to act on $G_3$ by exponentiation or $G_2$ by multiplication. In fact, we have to use the group of units of the ring $\Bbb{Z}$, which is isomorphic to $\mathbb{Z}_2$, so no progress can be made using this exact approach.2012-11-05
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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, 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.

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    @jak: Michael’s comment explains exactly why he’s talking about splitting things in half. A generator of$a$cyclic group cannot be split in half. Therefore, if $a$ is an element of some group $G$, and $a$ **can** be split in half, then $a$ cannot be$a$generator of $G$. If **everything** in $G$ can be split in half, then **nothing** in $G$ can be$a$generator of $G$, and therefore $G$ cannot be cyclic.2012-11-05
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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 $(a + a^2)/2$?