Let $G$ be a cyclic group of order $n$ and $K\leq AutG$ be a subgroup of the automorphism group of $G$. We denote the orbits of the natural action of $K$ on $G$ by $O\_1,\cdots, O\_s$. Let $\underline{X}\_i=\sum\_{x\in O\_i}x$ be the sum of elements in each orbit in the integral group ring $\mathbb{Z}G$. Then the $\mathbb{Z}$-span of $\underline{X}\_i$'s is a subring $\mathcal{A}$ of $\mathbb{Z}G$. Is $\mathcal{A}$ always generated by some $\underline{X}\_i$ as a ring over $\mathbb{Z}$?



generator of a subring of integral group ring
Not when $n=4$ and $G$ is the full automorphism group of $\mathbb Z/4$.


Then $\mathcal A$ is spanned by $\underline{X}\_1,\underline{X}\_2, \underline{X}\_4$ where $\underline{X}\_i$ is the sum of all elements of order $i$.


Then $\underline{X}\_1=1$ is the identity, $\underline{X}\_2^2 = 1$, $\underline{X}\_2 \underline{X}\_4 = \underline{X}\_4$, and $\underline{X}\_4^2 = 2 \underline{X}\_2 + 2$ so none can generate since $\underline{X}\_1=1$ simply generates $\mathbb Z$, $\underline{X}\_2$ generates the subspace spanned by $1$ and $\underline{X}\_2$, and $\underline{X}\_4$ generates the subspace spanned by $1, \underline{X}\_4, 2\underline{X}\_2$.


