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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?

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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.

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