Can we come up with a disjoint union of a subsets of the group $\Bbb{Z}$ such that they do not equal the cosets of a subgroup, yet they form a group?

Question

If this applies to $\Bbb{Z}$ it probably will work for other groups $G$, however, for simplicity and because I'm interested in integers & their primes, let's work with $G = \Bbb{Z}$.

Anyway, we all know that the cosets of $G$ can be added and subtracted elementwise. Meaning, although we define $(a\Bbb{Z} + x) + (a\Bbb{Z} + y) = a\Bbb{Z} + (x + y)$, you can also perform this addition with the middle $+$ on the left being an elementwise operation on the two sets (cosets), the result is the same.

So I'm wanting to partition $G$ into $G = A \uplus B \uplus \dots \uplus C$ in such a way that $A + B = C, \ B - C = D$, for example. The partition can be infinite or a finite number of classes $A, B, \dots$

In other words, the set $H = \{A,B, \dots\}$ forms a group under the elementwise operation, but there exists no ideal $n\Bbb{Z} \leqslant \Bbb{Z}$ such that $H = \Bbb{Z}/n\Bbb{Z}$. However, the group can and preferably should be isomorphic to one of $\Bbb{Z}$'s standard quotient groups.

Extra points if you can do this in terms of $\Bbb{P}$ the set of primes, and $\Bbb{C}$ the set of composite integers, or certain subsets thereof.

Can it be done?

Answer 1

It is not possible in any group. So, to be clear, we have a group $G$ and we are trying to come up wit a partition $G=\sqcup_i A_i$, where "termwise multiplication" yields to a group operation in the set $\{A_i\}_{i \in I}$.

Let us start by considering $E$ the class that contains the identity. Note that $E\cdot E=E$, as the left hand side contains $e^2=e$. Now, if $g, h \in E$, then $gh \in E\cdot E=E$. Also, if $g \in E$ and $B$ is the class that contain $g^{-1}$, then $EB$ contains $gg^{-1}=e$, hence $EB=E$. In particular, this shows that $B=E$, and so $g^{-1} \in E$. So far, $E$ is a group.

Now, if $h \in B$ is any class, let $C$ be the class containing $h^{-1}$. Then $BC$ contains $hh^{-1}=e$, so $BC=E$, and $C=B^{-1}$ in this group operation.

The same kind of argument shows that $E$ is a normal subgroup: if $h \in B$, then $BEB^{-1}$ contains $heh^{-1}=e$, hence $BEB^{-1}=E$.

At last, let $h \in B$. We claim that $B=hE$. In fact, from $B^{-1}B=E$, we get that for any $h' \in B$ we have $h^{-1}h' \in E$, hence $h'=h(h^{-1}h') \in hE$. This shows $B \subseteq hE$; the other inclusion is clear from $BE = E$.