There is a somewhat famous example of group cohomology witnessing $\mathbb{Z}/100$ as an extension of $\mathbb{Z}/10$ by $\mathbb{Z}/10$, with the standard carry function $c$ as a 2-cocycle (cf. this paper)
Of course, this is not the only extension of $\mathbb{Z}/10$ by $\mathbb{Z}/10$, and indeed if we use $2c$ as our carry function (i.e. we carry 2 instead of 1 on each stage) we get another extension, this time isomorphic to $\mathbb{Z}/2 \oplus \mathbb{Z}/50$. With the "standard" carry function $c$, we can iterate this process:
$\mathbb{Z}/1000$ is an extension of $\mathbb{Z}/100$ by $\mathbb{Z}/10$
$\mathbb{Z}/10,000$ is an extension of $\mathbb{Z}/1000$ by $\mathbb{Z}/10$
and so on.
It is clear that $\mathbb{Z}$ should be some sort of limit of these groups (I suspect it is the finitely supported subgroup of the inverse limit, but I haven't formally checked this), and I am curious how to extend this procedure to other groups and other carry functions.
Which brings us to my question:
If we take as our carrier set $(\mathbb{Z}/10)^*$ and define addition letter by letter, with the regular carry function $c$, we get $\mathbb{N}$. However, proving that this operation is associative requires knowledge of how the carries interact, which I'm having trouble adding to my induction.
As an example of where things become strange, consider 10 + 81 + 99, using $2c$ as our carry function:
associating left gives (10 + 81) + 99 = 91 + 99 = 400 (since 9+1=0, we carry a 2, 9+9=8, we carry a 2 for the 9+9 and a 2 for the new 8+2 from our old carry)
This issue of "carrying twice" doesn't come up in the standard carry case, because at any stage we have at most 9+9=18, with a carry is 19, and so we only ever carry 1 or 0. I'm not sure how to handle this in my induction.
Answers which generalize to cocycles over other groups are appreciated, as that is where I plan to take this.
I'm putting this in the automata theory tag as well, because they might be more familiar with methods of proving these sorts of operations associative.