# $\mathbb{Z}$ has no torsion?

What does is mean to say that $\mathbb{Z}$ has no torsion?

This is an important fact for any course?

Thanks, I heard that in my field theory course, but I don't know what it is.

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Seen this? – Sasha Aug 28 '12 at 16:37

It means that $$\forall\,n,z\in\Bbb Z\,\,,\,n\neq 0\,\,,\,nz=0\Longrightarrow z=0$$ And yes, it is a rather important notion in group theory in general.

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In other words, there are no elements of finite order besides the identity. – Mikko Korhonen Aug 28 '12 at 16:49
Indeed, that's exactly the meaning. – DonAntonio Aug 28 '12 at 16:55
You use "0" (usually an additive identity) but also multiplicative notation; how can that hold for general groups? m.k.'s explanation works for me but I'm not sure how yours translates. – Eric Stucky Aug 28 '12 at 21:12
Anyone understanding the group operation defined on $\,\Bbb Z\,$ is expected to know that $$nz:=\stackrel{\text{n times}}{\overbrace{z+z+...+z}}$$ – DonAntonio Aug 29 '12 at 0:07

A group $G$ has torsion if there are elements $g\neq0\in G$ such that $ng=0$ for some $n\neq 0$ (depending on $g$). In $\mathbb Z$, we have that $ng\neq 0$ for all $n,g\neq 0$ and so $\mathbb Z$ is torsion free.

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@d555, you might want to know that the notion of torison is extremely important, for example if $A$ is finitely generated and abelian group, then it can be written as the direct sum of its torsion subgroup $T(A)$ and a torsion-free subgroup (but this is not true for all infinitely generated abelian groups). $T(A)$ is uniquely determined.

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Thanks, you're right. These fact is interesting – jonaprieto Aug 31 '12 at 1:56

In general, given an $R$-module $M$, an element $m \in M$ is called a torsion element if there exists some non zero $r \in R$ such that $rm=0$. Here I denote the set of torsion elements of $M$ is denoted $\text{Tor}(M)$, although I have also seen it denoted by $T(M)$. If $\text{Tor}(M)=0$, then $M$ is said to be torsion free.

For any ring $R$, one can think of $R$ as an $R$-module over itself, where scalar multiplication is simply the ring multiplication. In your case you are looking at the ring $\mathbb{Z}$, which is certainly a $\mathbb{Z}$-module. So the torsion elements of $\mathbb{Z}$ would be the set $\text{Tor}(\mathbb{Z})$. If $a \in \text{Tor}(\mathbb{Z})$, then there exists some $n\in \mathbb{Z}$ such that $na=0$. Since $\mathbb{Z}$ is an integral domain (the prototypical one in fact), the equation $na=0$ forces either $n=0$ or $a=0$. Thus $\text{Tor}(\mathbb{Z})=\{0\}$.

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