Let D={$n_{i}$} be a sequence of integers, $n_{i+1}$ is a multiple of $n_{i}$ ($\forall i$) and $n_{i} \to \infty$. Let us consider a group $H(D)\subset \mathbb{Z}_{n_{0}} \times \mathbb{Z}_{n_{1}}\times \dots$, defined in the following way $$H(D)=\{(r_{0},r_{1},\dots) : r_{i+1}\equiv r_{i} (mod \ m_{i})\}$$
I have two questions, the group operation is to sum term by term? And what is the minimal shift in $H(D)$ by the element $(1,1,\dots)$?
Article: link
Thanks a lot.
The group operation in the direct product $\mathbb Z_{n_0} \times \mathbb Z_{n_1} \times \cdots$ is indeed to sum term by term. As $H(D)$ is a subgroup of this product it uses the same group operation so you are correct, the group operation on $H(D)$ is to sum term by term.
By "minimal shift" I assume you're asking what the order of the element $(1, 1, \ldots)$ is, let me know if you meant something else. Anyway, $(1, 1, \ldots)$ has infinite order because for any $n \in \mathbb N$ we have $n\cdot(1, 1, \ldots) = (n, n, \ldots)$ and $(n, n, \ldots) \neq (0, 0, \ldots)$ because the limit of the $n_i$ is infinity so there is some $i$ such that $n_i > n$ and hence $n \neq 0 \pmod{n_i}$.