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!

  • $\begingroup$ It's certainly torsion, but it isn't infinite. Hint: consider taking an infinite product of copies of a finite torsion abelian group. $\endgroup$ Apr 5, 2015 at 18:39
  • $\begingroup$ Thank you! So, you mean $\mathbb{Z}_p^{\infty}$, where $p$ is a prime? $\endgroup$ Apr 5, 2015 at 18:48
  • $\begingroup$ 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. $\endgroup$
    – pjs36
    Apr 5, 2015 at 18:49
  • $\begingroup$ Yes @pjs36, I actually mean rotation with all rational degree in a clock. $\endgroup$ Apr 5, 2015 at 18:51
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    $\begingroup$ 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 $n \in \mathbb{Z}$. $\endgroup$ Apr 5, 2015 at 19:01

2 Answers 2


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 $n$.

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    $\begingroup$ This group can also be called the torsion subgroup of the circle group. $\endgroup$ Apr 14, 2017 at 5:58

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.

  • $\begingroup$ So you have reduced the problem to finding an abelian group with no torsion, not finitely generated and with no three rationally independent elements. $\endgroup$ Jun 22, 2015 at 3:38
  • $\begingroup$ @ 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. $\endgroup$
    – CAA
    Jun 22, 2015 at 3:43

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