Name that group! Generalization of $A(p)$ for abelian groups.

Question

Have you seen or heard of the groups $\mathcal{A}(n)$ or $A(n)$ (for any integer $n$) described below?

This is the well-known construction:
Let $A$ be an abelian group. Then $A(p)$ is a subgroup which is the set of all elements $x\in A$ such that $ord(x) = p^k \ $ for some $k\in \mathbb{N}$. This, for prime $p$ that is, is a known construction talked about in Lang's Algebra. And if $A$ is torsion it's isomorphic to a direct sum of its nonzero $A(p)$ subgroups. Knowing that, what about $A(n)$ for any integer $n$? Let's see...

Here's the new construction that I'm wondering about:
Define $\mathcal{A}(n)$ to be the set of all elements that have an order that is a divisor of $n$. Then $\mathcal{A}(n)$ is a group.

Proof: If $x \in \mathcal{A}(n)$, then $dx = 0 \implies 0 = d(-x). \ $ But $\forall m : mx = 0, ord(x) \mid m$, so $ord(-x) \mid d \mid n$, so $ord(-x) \mid n$. If $x, y \in \mathcal{A}(n)$, let $ord(x) = e, ord(y) = e$; let $c = lcm(d,e)$. Then $c(x+y) = 0$. But notice that any divisor of $lcd(d,e)$ divides $n$. So $x+y \in \mathcal{A}(n)$. QED.

Then the $A(p) = \lim_{k\rightarrow \infty}\mathcal{A}(p^k) = \cup_k \mathcal{A}(p^k). \ $ That works for prime $p$, but also for any integer $n$. I.o.w. $A(n)$ is also a group.

What do you guys think?

Answer 1

The $A(p)$ is the $p$-power torsion subgroup of $A$, and $A(n)$ is the direct sum of all subgroups $A(p)$ where $p$ runs through the prime factors of$~n$.

Answer 2

The subgroup you denote as $\mathcal{A}(n)$ - consisting of all the elements of $A$ whose order divides $n$ - is often denoted $A[n]$ and is known as the $n$-torsion subgroup of $A$. Note that this terminology does not coincide with the common use of "$p$-torsion" to mean "$p$-power torsion" when $p$ is a prime number. This terminology is, however, a correct usage of the general case where an element $m$ of an $R$-module $M$ is referred to as $r$-torsion when $rm=0$.

The subset consisting of all the elements of $A$ whose order is a power of $n$ is not a subgroup of $A$ unless $n$ is prime. Otherwise there is a power of $n$ with a divisor that is not a power of $n$, and hence an element of $A(n)$ with a multiple that is not an element of $A(n)$.

The subset consisting of all the elements of $A$ whose order is a divisor of a power of $n$ is just the span of the subgroups $A(p)$ as $p$ ranges over the divisors of $n$, or in other words, the subgroup $$\sum_{p\mid n}A(p)$$ As far as I know there is no separate name for this subgroup.