Torsion subgroup

The torsion subgroup of an abelian group consists of all elements of finite order. For a finitely generated abelian group, it is the finite direct summand complementary to the free abelian part.

For an abelian group $G$ and an integer $m\ge 1$, its $m$-torsion subgroup is $G[m]=\{x\in G:mx=0\}$, the kernel of multiplication by $m$. In a finite abelian group, every nonempty fiber of this homomorphism has cardinality $|G[m]|$, so $|G:mG|=|G[m]|$. This follows directly: if $mx=my$, then $x-y\in G[m]$, and conversely translating by an element of $G[m]$ leaves the image unchanged. On the cyclic group ${\mathbb{Z}}_N$, the solutions of $mx=0$ number $\gcd (m,N)$; divide both integers by their greatest common divisor to see that $x$ must be a multiple of $N/\gcd (m,N)$.

For an abelian group $A$, its $p$-primary torsion subgroup is $A\{p\}=\{a:p^{r}a=0\text{ for some }r\ge 0\}$. It is zero exactly when $A[p]=0$, since a nonzero element of $p$-power order has a multiple of order $p$. Multiplication by an integer prime to $p$ is an automorphism on it.