Romanian Mathematical Olympiad - Final Round

a. We will show that the group $(\mathbb{Z}_n, +)$ is arrangeable if and only if $n \ge 2$ is an odd positive integer.

If $(G, \cdot)$ is an abelian arrangeable group, then considering the arrangement $G = \{a_1, a_2, \dots, a_k, \dots, a_n\} = \{a_1 \cdot a_2, a_2 \cdot a_3, \dots, a_k \cdot a_{k+1}, \dots, a_n \cdot a_1\}$, we have
$$\prod_{g \in G} g = \prod_{k=1}^{n} a_k = \prod_{k=1}^{n} (a_k \cdot a_{k+1}) = \left( \prod_{g \in G} g \right)^2,$$
(where $a_{n+1} = a_1$), so that $\prod_{g \in G} g = 1$, where $1$ is the unit element of the group $(G, \cdot)$.

In any finite abelian group the product of all the elements is equal to the product of all its elements of order $2$.

For $n \in \mathbb{N}^*$, $n \ge 2$, if $k \in \{0, 1, \dots, n-1\}$ with $\mathrm{ord}(\hat{k}) = 2$, then
$$\hat{k} \neq \hat{0} = \hat{k} + \hat{k} = 2\hat{k},$$
so that $n$ divides $2k$, but does not divide $k$. This is only possible if $n$ is even and $n = 2k$. Thus, if $n$ is even, with $n = 2k$, and $(\mathbb{Z}_n, +)$ were arrangeable, we would have
$$\hat{0} = \sum_{x \in \mathbb{Z}_n} x = \hat{k},$$
which is false. Hence, if $n$ is even, the group $(\mathbb{Z}_n, +)$ is not arrangeable.

Let now $n \ge 3$ be an odd positive integer. We consider the set $[0, n-1]_N = \{0, 1, \dots, n-1\}$ and the function $f : [0, n-1]_N \to \mathbb{Z}_n$, defined by $f(k) = \overline{2k+1}$.

Since
$$f(k) = f(l) \iff \overline{2k+1} = \overline{2l+1} \iff n|(2k-2l) \iff n|(k-l) \iff k=l,$$
the function $f$ is injective and since $[0, n-1]_N$ and $\mathbb{Z}_n$ are finite sets with equal cardinals, it follows that $f$ is bijective. Denoting $a_k = \overline{k-1}$ for any $1 \le k \le n$ and $a_{n+1} = a_1$, it follows then that $a_k + a_{k+1} = f(k-1)$ for any $k = \overline{1, n}$, so that $\mathbb{Z}_n = \{a_1, a_2, \dots, a_n\} = \{f(0), f(1), \dots, f(k), \dots, f(n-1)\} = \{a_1 + a_2, a_2 + a_3, \dots, a_{k-1} + a_k, \dots, a_n + a_1\}$, whence we deduce that the group $(\mathbb{Z}_n, +)$ is arrangeable.

The set of all positive integers such that the group $(\mathbb{Z}_n, +)$ is arrangeable is thus the set of all odd positive integers $n$, with $n \ge 3$.

b. According to part a), there are no cyclic arrangeable groups of even order. We consider $\mathbb{Z}_4 = \{\hat{0}, \hat{1}, \hat{2}, \hat{3}\}$, $\mathbb{Z}_2 = \{\overline{0}, \overline{1}\}$ and the group $G = \mathbb{Z}_4 \times \mathbb{Z}_2$ with component-wise defined addition $(\hat{k}, \hat{l}) + (\hat{m}, \hat{n}) = (\overline{k+m}, \overline{l+n})$. Then $G = \{a_1 = (\hat{0}, \overline{0}), a_2 = (\hat{1}, \bar{0}), a_3 = (\hat{1}, \bar{1}), a_4 = (\hat{3}, \bar{1}), a_5 = (\hat{2}, \bar{0}), a_6 = (\hat{2}, \bar{1}), a_7 = (\bar{0}, \bar{1}), a_8 = (\hat{3}, \bar{0})\} = \{a_1+a_2, a_2+a_3, a_3+a_4, a_4+a_5, a_5+a_6, a_6+a_7, a_7+a_8, a_8+a_1\}$, so that $(G, +)$ is an arrangeable group of order $8$.