What is $mA$, where $m \in \mathbb N$ and $A$ is a group?

Question

Simple question. I was reading through a proof of Mordell's Theorem here, and in the statement of the Descent Theorem (page 4), the notation $A/mA$ is used, where $m \geq 2$ is an integer and $A$ is an abelian group. I have completely forgotten what $mA$ means in this case. I figure it is $$mA = \{m \cdot x : x \in A\},$$ where $m \cdot x = \underbrace{x + x + \ldots + x}_{m\text{ times}}.$ Is this correct? (It certainly isn't $mA = \underbrace{A + A + \ldots + A}_{m\text{ times}}$, because then we would have $mA = A$.)

Answer 1

Yes, your interpretation of it is correct. The notation is quite common when dealing with an Abelian group written additively; the most familiar example is probably $m\Bbb Z$, the subgroup of $\Bbb Z$ consisting of the multiples of $m$.

Answer 2

The notation $A/nA$ is indeed quite common, and arises very often in the context of abelian groups. You gave an example in number theory, because the group of an elliptic curve over $\mathbb{Q}$ is finitely-generated abelian; another context is in homological algebra. For example, for every abelian group $A$ we have $\mathbb{Z}/n\otimes_{\mathbb{Z}}A\cong A/nA$. This is important for the Tor functor. But, this is only a comment. The answer to your question is simply "yes".