My book defines an R-module as an abelian group X with the defined operation:
$$\phi:R\times X \to X \\\phi(x) = \alpha x$$
Such that these conditions are satisfied:
$$\alpha(x+y) = \alpha x + \alpha y$$
$$(\alpha+\beta)x = \alpha x + \beta x$$
$$\alpha(\beta x) = (\alpha\beta) x$$
$$1x = x$$
It then defines a submodule $A$ of the $R$-module $X$ as a subset of $X$ that is also a module over $R$, that is, by definition, $A$ is a subgroup of the abelian group $X$ and it must be closed under the operation $\alpha x$, for $x\in A$
Then it introduces this lemma, which is what I need to know if I understood
A nonempty subset $A$ of module $X$ over $R$ is a submodule of $X$ $\iff$, for arbitrary elements of $\alpha\in R$ and $u,v\in A$ we have $u+v\in A$ and $\alpha u \in A$
In the proof of this lemma, it says that by the definition of submodules, the condition is evidently necessary. It remains to estabilish the sufficiency of the condition. For this purpose, it suffices to prove that the negative $-u$ of every element $u\in A$ is in $A$. This follows from:
$$-u = (-1)u\in A$$
I didn't understand which things are satisfied by this. I can see that now we're dealing with a subset, not a subgroup, therefore we need to check that this forms a group. I can see, at least, that the condition $u+v\in A$ is the closure definition of a group. Since it's a subset of a module, the associative operations are still valid, and the identity is valid as well. It suffices to prove that the inverse it's in $A$, which is what he did.
The condition $\alpha x \in A$ is just for the submodule to be well defined, I guess