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I am new to the concept of torsion. Is there any example for an infinite torsion abelian group?

Here is my example: rotation with a rational degree in a clock. Is this an example?

Thank you very much!

breezeintopl
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  • It's certainly torsion, but it isn't infinite. Hint: consider taking an infinite product of copies of a finite torsion abelian group. – Viktor Vaughn Apr 05 '15 at 18:39
  • Thank you! So, you mean $\mathbb{Z}_p^{\infty}$, where $p$ is a prime? – breezeintopl Apr 05 '15 at 18:48
  • I'm not positive, but I think by "rotation with rational degree", they meant the set $\left{\left(\frac{a}{b}\right)\cdot 1^{\circ} : a,b \in \Bbb Z\right}$ of rotations not just for a fixed rational number, but consisting of all rational multiples of $1$ degree. – pjs36 Apr 05 '15 at 18:49
  • Yes @pjs36, I actually mean rotation with all rational degree in a clock. – breezeintopl Apr 05 '15 at 18:51
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    Ah, I see you didn't mean for some particular rational degree--you meant all rational degrees! Yes, then your group is isomorphic to $\mathbb{Q}/\mathbb{Z}$ (which is isomorphic to pjs36's answer). But yes, I was suggesting $G = \prod_{i=1}^\infty \mathbb{Z}/p\mathbb{Z}$. Your answer is actually more interesting: while $p \cdot G = {0}$ for my example, $n \cdot \mathbb{Q}/\mathbb{Z} \neq {0}$ for all $n \in \mathbb{Z}$. – Viktor Vaughn Apr 05 '15 at 19:01
  • Thank you all for your answers! – breezeintopl Apr 05 '15 at 19:02

2 Answers2

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Sufficiently clarified, your example works.

It's isomorphic to the multiplicative group of complex numbers $\{z \in \Bbb C : z^n = 1 \text{ for some }n \in \Bbb Z\}$ that have a finite multiplicative order; the union of $n$th roots of unity over all integer $n$.

pjs36
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Let $G$ be an abelian group having no torsion elements, not finitely generated and let $G$ has only two rationally independent elements. Let $a$ and $b$ be those two elements. This means if $xa + yb = 0,$ then $x = 0$ and $y = 0,$ where $x$ and $y$ are integers. You can think $x$ and $y$ are also rational numbers and then you can clear the denominator in $xa + yb = 0$. Let $H$ be the subgroup of $G$ generated by $a$ and $b.$ Then the group $G/H$ is an infinite torsion group. Since $G/H$ has no rationally independent elements so it is a torsion group, and $G$ is not finitely generate and $H$ is finitely generated implies $G/H$ is an infinite group.

CAA
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  • So you have reduced the problem to finding an abelian group with no torsion, not finitely generated and with no three rationally independent elements. – Mariano Suárez-Álvarez Jun 22 '15 at 03:38
  • @ Mariano Suárez-Alvarez Thank you. Actually I didnot think that way but as an example we can go more simple form. I think we can just take $G = \mathbb Q, H = \mathbb Z.$ Then $\mathbb Q/\mathbb Z$ is an infinite torsion group. – CAA Jun 22 '15 at 03:43