Correction to: On subset sums of $\mathbb {Z}_n^{\times }$ which are equally distributed modulo n

Correction to: Arch. Math. https://doi.org/10.1007/s00013-023-01853-2

It has been brought to the author’s attention that the statement of [1, Theorem 2.4] is incorrect. However, it is easily corrected by adding the following relation to its statement: The number r is the minimum number such that

$$\begin{aligned} 2^r \equiv 1\pmod {q} \quad \text {or}\quad 2^r\equiv -1\pmod {q}. \end{aligned}$$

The reason is the following:

In the proof, we deduce that there is a permutation s of $\{1, 2,\ldots , k\}$ such that $\langle 2a_i\rangle =\langle a_{s(i)}\rangle$. We need to show that $\langle 2a_r\rangle =\langle a_1\rangle$ in order to define $B_1=\{a_1,\ldots ,a_r\}=\{b_1\cdot (\pm 2^{j-1})\}$ with leader $b_1=a_1$.

But $\langle 2a_r\rangle =\langle a_1\rangle$ holds if and only if $a_1\equiv \pm 2a_r$, which is equivalent to $a_1\equiv \pm 2^ra_1\pmod {q}$. Since $\gcd (a_1, q)=1$, this is possible if $2^r\equiv \pm 1 \pmod {q}$. The careful reader may observe that $2^r\equiv -1\pmod {q}$ is possible only if the order of 2 modulo q is even. This means that if the order of 2 modulo q is odd, the statement is valid without any corrections.

In addition, at the beginning of page 6, the phrase

“From this construction, it is also evident that $a_1\equiv \pm 2a_r$ since $2^r\equiv 1\pmod {q}$” should be

“From this construction, it is also evident that $a_1\equiv \pm 2a_r$ since $2^r\equiv \pm 1\pmod {q}$”.