Is $\mathbb{Z}/n\mathbb{Z}$ a group under multiplication when $n$ is prime?

Question

I have to prove that $\mathbb{Z}/n\mathbb{Z}$ is not a group under multiplication for all n>1. I would argue, however, that when $n$ is prime, $\mathbb{Z}/n\mathbb{Z}$ is a group under multiplication. Multiplication is associative, every element in $\mathbb{Z}/n\mathbb{Z}$ has an inverse when $n$ is prime, and the identity element is simply $\bar{1}$. What am I missing?

Answer 1

For $\mathbb{Z}/n\mathbb{Z}$ to be a group, we need that for every $a \in \mathbb{Z}/n\mathbb{Z}$, to exist $b \in \mathbb{Z}/n\mathbb{Z}$ such that $a*b = 1$. But for $a = 0$, there is no $b$ such that $a*b = 1$, because $a*b = 0, \forall b \in \mathbb{Z}/n\mathbb{Z}$. Then, $\mathbb{Z}/n\mathbb{Z}$ is not a group!

Answer 2

Maybe it worth asking when is $(Z/nZ)^*$ a group under multiplication and the answer is precisely when $n$ is prime. Otherwise, $n=ab$ would imply that the product of the residue classes of $a$ and $b$ does not belong to $(Z/nZ)^*$