Trying to understand what this ring notation means

Question

I saw $(\Bbb Z/3\Bbb Z - \{0\}, \cdot)$. What does the $-\{0\}$ do?

Answer 1

This is the group obtained from

• the field $\mathbb{Z}/3\mathbb{Z}$ (the integers modulo 3),
• without the additive identity, $0$,
• under the operation of multiplication.

The result has two elements $1 \pmod{3}$ and $2 \pmod{3}$, with the usual multiplication modulo $3$, forming a group, the group of units modulo $3$.

If we leave the additive identity in, we do not get a group, because $0$ has no multiplicative inverse. (Units are the elements that have multiplicative inverses.) We get a semigroup, but groups are cool (much like (unrelated) fezzes and bowties). More generally, we would want to exclude all non-invertible elements. If we were starting with $\mathbb{Z}/10\mathbb{Z}$, we would exclude $0$, $2$, $4$, $5$, $6$, and $8$ modulo $10$, so that the remaining elements can form a group, the group of units modulo $10$.

Answer 2

That is the set difference operator. Here, $\mathbb{Z}_{3} - \{0\} = \{ 1, 2\}$. So the multiplicative group for $\mathbb{Z}_{3}$ contains the non-zero elements of $\mathbb{Z}_{3}$.