The Making of a Ring — How can we choose 1? I am an undergraduate studying Algebra: Chapter 0 by Paolo Aluffi in my free time (Note that because of this whenever I refer to 'rings' I am speaking of 'rings with identity'). While learning about the basics of rings, it became clear to me that we define multiplication by associativity and distributivity because these are the properties of group homomorphisms. Ring multiplication distributes over addition in the underlying abelian group because of the homomorphism condition—i.e. for a group homomorphism $\varphi: G \to G$, we have $\varphi(a+b) = \varphi(a) + \varphi(b)$. Similarly the associativity of multiplication is due to the associative nature of morphism composition.
Therefore it seems natural to me to view the multiplication operation as a convenient abstraction encapsulating the data we obtain by studying an abelian group alongside its group endomorphisms. Aluffi brings as Proposition 2.7 the following:

Let $R$ be a ring. Then the function $r \mapsto \lambda_r$ [where $\lambda_r$ is the group endomorphism representing left ring multiplication by $r$] is an injective ring homomorphism
$$\lambda: R \to \mathrm{End}_{\mathrm{Ab}}(R).$$
(Algebra: Chapter 0, §III.2.5)

Thus the image of $\lambda$ is a subring of $\mathrm{End}_{\mathrm{Ab}}(R)$ isomporphic to $R$. If we can pick $\mathrm{im} \ \lambda$ out of $\mathrm{End}_{\mathrm{Ab}}(R)$ without a predefined ring structure, we can obtain a ring structure on $R$ using an isomorphism $\mathrm{im} \ \lambda \to R$ given by $\lambda_r \mapsto \lambda_r(1_R)$. So given an abelian group $G$ and an element designated $``1_G"$, I think that we need only to study how a certain subgroup of $\mathrm{End}_{\mathrm{Ab}}(G)$ interacts with $1_G$ to determine a ring structure on $G$. Therefore my question is: What are the precise consequences of differing choices of $1$ when defining a ring structure on an abelian group?
My thoughts on this so far:
Clearly different choices of $1$ lead to different ring structures. If, for example, we let $G := (\mathbb{Z}\backslash 4\mathbb{Z}, +)$, the natural choice of $1_G$ being 1 gives rise to a ring of characteristic $4$. Now, I struggled to actually carry out this example, but we can easily imagine that if we could choose $1_G$ to be 2, then the induced ring would be of characteristic 2, since $2 + 2 = 2 \cdot 1_G = 0$.
Furthermore, since 3 is the additive inverse of 1 in this group, we should be able to let $1_G = 3$ and come up with a sort of 'backwards' version of the usual ring structure. In this case we have a ring with $G$ as the underlying abelian group, isomorphic to the version with $1_G = 1$ and multiplication defined by:




$\ast$
3
2
1




3
3
2
1


2
2
0
2


1
1
2
3




So is every element of an arbitrary abelian group $G$ a candidate for $1_G$ when defining a ring structure? If so, what are the consequences of different choices, and if not what are the necessary group-theoretic requirements?
Note: I am a beginner at this, having only studied this subject independently throughout my senior year of highschool and freshman year of college. With that in mind please let me know if anything I said is off-base and try to explain things as simply as possible. Thank you.
 A: Below, "ring" means "unital ring."
In the comments above, runway44 has pointed out that there is in fact a huge gap between a full ring structure on $G$ and a choice of multiplicative identity. However, I'm going to focus on the question you've asked specifically:

is every element of an arbitrary abelian group $G$ a candidate for $1_G$ when defining a ring structure? If so, what are the consequences of different choices, and if not what are the necessary group-theoretic requirements?

Given an abelian group $G$, let $I(G)$ be the set of elements $a$ of $G$ such that there is a ring $R$ with additive group $G$ and multiplicative identity $a$. At a most basic level, we just want to compute $I(G)$ given $G$.
Let's start with a triviality. If $G$ is nontrivial then $I(G)$ is not all of $G$ since the additive identity is also a multiplicative annihilator in any ring. Additionally, there are some "un-ringable" abelian groups - that is, nontrivial groups satisfying $I(G)=\emptyset$. The simplest example of such in my opinion is $\mathbb{Q}/\mathbb{Z}$.
Now we come to a slightly less trivial observation: we can in fact have $\emptyset\subsetneq I(G)\subsetneq G\setminus\{e_G\}$. Given a ring $R$ and an element $r\in R$, let $ord_R(r)$ be the order of the cyclic subgroup of the additive group of $R$ generated by $r$. For example, if $R$ is a field and $r$ is the multiplicative unit, then $ord_R(r)$ is just the characteristic. Distributivity implies that $ord_R(r)\vert ord_R(1_R)$ for every $r\in R$. Turning this around, we get a constraint on the function $I$ defined above: for example, every element of $I(\mathbb{Z}/6\mathbb{Z})$ must have order $6$ (and since all elements of order $6$ in $\mathbb{Z}/6\mathbb{Z}$ are automorphic, this gives a complete description of $I(\mathbb{Z}/6\mathbb{Z})$).
A: I want to make a philosophical or maybe pedagogical point that it is almost never the case in ring theory that we define a ring by starting with an abelian group and working hard to figure out a multiplication to put on it; that's not really how rings arise in practice at all. In practice the multiplication is usually completely obvious and a key part of why we wanted to study the ring in the first place. Noncommutative rings typically arise as rings of endomorphisms on some abelian group (in which case the identity is the identity endomorphism and multiplication is composition) and commutative rings typically arise as rings of functions in some sense, e.g. polynomials or continuous functions (in which case the identity is the function with constant value $1$ and multiplication is pointwise multiplication).
The main counterexample I know is deformation quantization which is definitely not an introductory or mainstream topic in ring theory. In deformation quantization the work really is defining a multiplication. Identities play a relatively minor role here still, though.
