# Cyclic group generator and multiplicative identity of correspondng ring

Can cyclic groups made into ring with unity such that multiplicative identity is not any generator? (Or does there exist example of one such cyclic group?)

Can we make $(\mathbb{Z}, +)$ into ring with unity without $+1$ or $-1$ (belonging to $\mathbb{Z}$) as multiplicative identiy.

Motivation: In rings with unity, characteristic of ring is order of multiplicative identity. Does it has some relation with generator of additive group if it is cyclic?

For $R = \mathbb Z$, let $\times$ denote the usual multiplication on $\mathbb Z$, and $\odot$ some other multiplication. Suppose $\odot$ has multiplicative identity $k$. Then
\begin{align} 1 = k\odot 1 &= \overbrace{(1+\cdots + 1)}^{k \text{ times}}\odot 1\\ & = \overbrace{1\odot 1 + \cdots + 1\odot 1}^{k \text{ times}} &&\text{by distributivity}\\ &= k\times(1\odot 1)&&\text{by definition} \end{align}
So we must have $1\odot 1 = \frac 1k$, whence $k = \pm 1$, since $\frac 1k$ must be an integer.
For $R = \mathbb Z/n\mathbb Z$, the same argument shows that
$$k\times(1\odot 1) \equiv 1 \pmod n$$ so $k$ must be invertible modulo $n$. So $k$ is coprime to $n$, and hence it generates the cyclic group.