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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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Transcribed image to text...I hope I didn't miss anything! – amWhy Nov 5 '12 at 0:07
@amWhy: \mid gives better spacing than a plain pipe. – Brian M. Scott Nov 5 '12 at 0:48
@Brian: thanks for the tip! I've actually learned everything I know about TeX from editing/answering questions here on looking for more efficient techniques than brute force! I'm not very fast, so I miss out on answering a lot of questions (when others manage to answer quickly and well). – amWhy Nov 5 '12 at 0:53
@amWhy: I’ve also learned 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. – Brian M. Scott Nov 5 '12 at 0:57

3 Answers 3

up vote 3 down vote accepted

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 $(a + a^2)/2$?

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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 $0<q\in\Bbb Q$, 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.

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Isn't there a way to tell by just quick inspection without doing this kind of work? – Hawk Nov 5 '12 at 0:00
@jak: This is really very little work, especially for $G_2$. Many of the proofs and problems that you’ll encounter in abstract algebra require much more effort. – Brian M. Scott Nov 5 '12 at 0:08
@jak : Your comment makes me wonder if you know what "cyclic" means. A cyclic group looks like this: $\{0, \pm a,\pm2a,\pm3a,\ldots\}$. It may go on forever, or it may return to $0$ after finitely many steps. If there's nothing that could be $a$, then the group is not cyclic. For example, if $a$ can be split in half, so that $b+b=a$, then $a$ does not fit in the role of a generator. In $(\mathbb Q, +)$, everything can be split in half, so nothing can be a generator, so it's not cyclic. – Michael Hardy Nov 5 '12 at 0:37
@MichaelHardy, why are you talking about splitting things in half? – Hawk Nov 5 '12 at 1:18
@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. – Brian M. Scott Nov 5 '12 at 1:32

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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Do you mean that $\phi$ is defined by $\phi(n)(x)=x^n$? But then $\phi(2)$ is not an automorphism of $\langle\mathbb Q^+,\cdot\rangle$ as it is not surjective. I like this approach but either I'm missing something or it doesn't work in this case. – nonpop Nov 5 '12 at 8:32
@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. – peoplepower Nov 5 '12 at 8:41

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