Can $\mathbb{Z}/n\mathbb{Z}$ (not $(\mathbb{Z}/n\mathbb{Z})^{\times}$) be a group under multiplication?

Question

I was wondering why we usually say $\mathbb{Z}/n\mathbb{Z}$ is a group under addition and invent notation like $(\mathbb{Z}/n\mathbb{Z})^\times$ specifically for the multiplicative group modulo $n$. Can $\mathbb{Z}/n\mathbb{Z}$ (not $(\mathbb{Z}/n\mathbb{Z})^\times$) be a group under multiplication?

Answer 1

Presuming that $1 \not\equiv 0 \mod n$, i.e., $n>1$, what could the multiplicative inverse of $0$ be?

The existence of an inverse for each element is the only axiom of a group that is failed by $(\mathbb Z/n\mathbb Z, \cdot)$. Accordingly, in order to form a group, $(\mathbb Z/n\mathbb Z)^\times$ consists of precisely those elements of $\mathbb Z/n\mathbb Z$ which have a multiplicative inverse.

Answer 2

We often refer to an abelian group operation as $+$, probably in part because many common abelian groups ($\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}, \mathbb{Z} / n \mathbb{Z}, \mathbb{F}[x]$, etc.) are underlying groups (under addition) of common ring structures. This becomes slightly peculiar when (as we often do) we use "multiplication" as a synonym for the group operation of a general group.

Answer 3

${\mathbb Z}_n$ is a monoid (a semigroup with 1) under multiplication. 0 is its absorbing element. Absorbing element can’t be invertible unless the monoid is trivial (1 = 0); that’s why for $n>1$ there is no group.