Infinite cyclic group only has two generators

[Mr Davis 97]

Homework Statement

Let $H = \langle x \rangle$. Assume $|x| = \infty$. Show that if $H = \langle x^a \rangle$ then $a = \pm 1$

Homework Equations

The Attempt at a Solution

Here is my attempt: Suppose that $H = \langle x^a \rangle$. Then, for arbitrary $b \in \mathbb{Z}$, $x^b = (x^a)^m$ for some $m \in \mathbb{Z}$. Then $x^{b-ma} = 1$, which is only true when $b = ma$. But for the last equation $b$ is arbitrary, so there only exists an $m$ for arbitrary $b$ such that $b = am$ when $a = \pm 1$. To the contrary, suppose that $a \ne \pm 1$ and $b \ne a$; then we can choose $b$ to be relatively prime with $a$ so that there exists no $m$ s.t. $b = am$.

Is this clear and rigorous? I feel that there could be a better way of showing that $b = am \implies a = \pm 1$.

 

[haruspex]

which is only true when b = ma.

That is not so in general. Why is it so in this case, though?

 

[fresh_42]

Homework Statement

Let $H = \langle x \rangle$. Assume $|x| = \infty$. Show that if $H = \langle x^a \rangle$ then $a = \pm 1$

Homework Equations

The Attempt at a Solution

Here is my attempt: Suppose that $H = \langle x^a \rangle$. Then, for arbitrary $b \in \mathbb{Z}$, $x^b = (x^a)^m$ for some $m \in \mathbb{Z}$. Then $x^{b-ma} = 1$, which is only true when $b = ma$. But for the last equation $b$ is arbitrary, so there only exists an $m$ for arbitrary $b$ such that $b = am$ when $a = \pm 1$. To the contrary, suppose that $a \ne \pm 1$ and $b \ne a$; then we can choose $b$ to be relatively prime with $a$ so that there exists no $m$ s.t. $b = am$.

Is this clear and rigorous? I feel that there could be a better way of showing that $b = am \implies a = \pm 1$.

No. That's a good example why I favor to write all dependencies explicitly. E.g. people write "$\forall \, \varepsilon \,\exists \,N$ such that ..." Wrong! It has to be "$\forall \, \varepsilon \,\exists \,N(\varepsilon)$ such that ..." because the choice of $N$ depends on the choice of $\varepsilon \,.$

It is the same in what you wrote: $\forall \,b\in \mathbb{Z} \,\exists \, m(b)\in \mathbb{Z} \ldots$ We thus have $b=m(b)\cdot a$. So what? This is no surprise as it is exactly what we started with.

There is an easy way to check your proof, because $(H,\cdot)=(\mathbb{Z},+)$. The exponents of the multiplicative $x$ are precisely the integers, so that is the isomorphism. Just assume $a \neq \pm 1\,.$ How could we ever reach points between $1$ and $a\,?$ Only if we had a relation, but $x$ doesn't have a relation, it is a free group.

Edit: Correction. Now I got your argument. $a \,\mid \,b$ for all $b$ and thus $a$ is a unit, hence $a= \pm 1\,.$ Sorry. It's a bit late here.

 

[Mr Davis 97]

That is not so in general. Why is it so in this case, though?

It is true in this case because $|x| = \infty$. If I add this in would the proof be fine? When I read it's a bit hard to follow (as evidenced by fresh_42's edit), so O just want to make sure there's not some clearer way that uses the same underlying logic.

 

[haruspex]

It is true in this case because |x|=∞.

Yes, but that step bears a little more detail.