Can we make an abelian group into a ring by defining multiplication only on a generating set? Suppose we have an abelian group $(G,+)$ that is generated by some set $A\subseteq G$. Suppose that we are able to define a binary operation $\ast$ on $A$, i.e.
$$\ast:A\times A\to A,\quad (a,b)\mapsto a\ast b,$$
such that $\ast$ is commutative, associative and such that there is an element $1\in A$ with the property that $1\ast a=a$, $\forall a\in A$.
Can we always make $G$ into a ring $(G,+,\ast)$ by just "imposing" the distributivity laws? That is, we just define, for example, $a\ast(b+c):=a\ast b+a\ast c$ for $a,b,c\in A$, and so on. Is this always well-defined? If not, what do we need to check to say it is well-defined? Are there additional properties that $\ast$ has to satisfy?
 A: If $G = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, with generators $1$ and $a$ of infinite order and $b$ of order 2, define $\ast$ by $a \ast a = a \ast b = b \ast b = a$, and $1 \ast a = a$, $1 \ast b = b$, $1 \ast 1 = 1$. This is associative and commutative with 1, but:
$a \ast (b + b)  = a \ast b + a \ast b = a + a = 2a \neq 0$
But $b + b = 0$, so we would have to have $a \ast (b + b) = 0$.
If $G$ is freely generated by $A$ it seems like it does work, however.
A: The short answer is: possibly, but you need an additional assumption to ensure that it works.
You've identified an associative binary operation $*$ on the set $A$ with respect to which $A$ is closed and has an identity element. Another way to say this is that $(A,*)$ is a monoid.
Now at the same time, your group $(G,+)$ consists of finite $\mathbb{Z}$-linear combinations of elements from $A$, i.e., every $g \in G$ can be written as a sum $g = \sum_{a \in A}z_a a$ where $z_a \in \mathbb{Z}$ for all $a \in A$. Given a second element $g' = \sum_{a \in A}z_a' a$ of $G$, you want to define the product $g*g'$ in such a way that it's compatible with its definition on $A\times A$. We can do this by defining $g * g'$ as the convolution
$$
g*g' = \left(\sum_{a\in A} z_{a} a\right)\left(\sum_{b\in A} z_{b}'b\right)
    = \sum_{c \in A}\sum_{a*b = c}z_{a}z_{b}' c
$$
(Note that this is equivalent to imposing a distributive law.)
So, yes, there might be a way to extend $*$ to an operation on $G$, but, as you noticed, it isn't clear that $g * g'$ is well-defined. Specifically, the problem is that $G$ isn't necessarily freely generated by $A$ given your assumptions, which is to say that there might not be a unique way to express $g \in G$ as a linear combination of elements of $A$. If $(G,+)$ is a free group on $A$, then $(G,+,*)$ is a ring and, in fact, isomorphic to the monoid ring $\mathbb{Z}[A]$. Otherwise you would need to verify the operation is well-defined by verifying that if $\sum y_a a = \sum z_a a$ and $\sum y_a' a = \sum z_a a'$ then
$$
    \left(\sum y_a a\right)\left(\sum y_a' a\right) = \left(\sum z_a a\right)\left(\sum z_a' a\right).
$$
Once you know that to be true, then $(G,+,*)$ is again a ring.
A: If, instead of "making $G$ into a ring" as you ask, you're satisfied with "building a ring that contains $G$", you can look into the group ring concept. However, instead of giving the ring's addition, $G$ will give the ring's multiplication by writing $G$ multiplicatively and extending linearly.
The upshot is this works for all groups, rather than only groups which are free over the a generating set.
