1

Consider the group $\bigoplus_{i=-\infty}^{\infty}\mathbb{Z}_{(i)}$, where each $\mathbb{Z}_{(i)}$ is a copy of $\mathbb{Z}$. Given $x \in \bigoplus_{i=-\infty}^{\infty}\mathbb{Z}_{(i)}$, let $x^{(k)}$ be the element in $\bigoplus_{i=-\infty}^{\infty}\mathbb{Z}_{(i)}$ obtained by shifting every entry of $x$ to the right by $k$ places. Let $x_1, x_2, \ldots, x_n \in \bigoplus_{i=-\infty}^{\infty}\mathbb{Z}_{(i)}$ be some non-zero elements.

I was wondering if there is a bound for the number of $x_j$ we can have such that

  1. the subgroup $ H = \langle x_j^{(i)} \mid i \in \mathbb{Z}, 1 \leq j \leq n \rangle $ is a proper subgroup
  2. $H \neq \langle x_j^{(i)} \mid i \in \mathbb{Z}, 1 \leq j \leq n, j \neq k \rangle $ for $1 \leq k \leq n \rangle$, i.e. if we remove one of $x_j$'s, we get a strickly smaller subgroup.

I was looking at this example: let $x_1 = (\ldots, 0, 1, 1, 0, \ldots)$ and $x_2 = (\ldots, 0, 2, 0 , \ldots)$. I don't seem to be able to find another element $x_3$ that satisfies the above condition.

(This is a follow-up question to the one I asked here.)

ghc1997
  • 2,077
  • 1
    But, if you take $\delta_0$ the kronecker delta function at $0$, don't you get $H=\bigoplus\mathbb{Z}$? my point is, why do you expect that $H$ can be proper if you put a bound on the number of elements? – Just dropped in Jun 23 '23 at 21:00
  • 1
    I don't understand. What is $k$? Do you mean for each $k = 1, \dots, n$? If so you could take each $x_j = n_j \delta_0$ where $n_j = 2 p_1 \cdots p_n / p_j$ and $p_1, \dots, p_n$ are the first $n$ odd primes. – Sean Eberhard Jun 27 '23 at 08:40

0 Answers0