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

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?

• $0$ hasn't got an inverse.
– lulu
Commented May 13, 2016 at 17:50
• oh. So then isn't it true that $\mathbb{Z}/n\mathbb{Z}$ is not a group under multiplication for all $n$? Commented May 13, 2016 at 17:51
• As lulu noted, $0$ has no inverse. Though, $\Bbb Z/p\Bbb Z \setminus \{0\}$ is a group
– user258700
Commented May 13, 2016 at 17:51
• @Obliv That's a lot of negatives. $\mathbb Z / n\mathbb Z$ is never a group under multiplication because the element $0$ is never invertible.
– lulu
Commented May 13, 2016 at 17:52
• how do I prove,then, that $\mathbb{Z}/n\mathbb{Z}$ is not a group under multiplication? Do I simply show how 0 is not invertible and that 0 is in $\mathbb{Z}/n\mathbb{Z}$ for all n>1? Commented May 13, 2016 at 17:54

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!
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 $(Z/nZ)^*$
• yeah but that doesn't change the fact that $0$ is in $\mathbb{Z}/n\mathbb{Z}$ regardless of what $n$ is. It'll never be a group as long as $0$ is in it. Commented May 13, 2016 at 18:06
• True. That's why the question is not interesting. But the absence of $0$ makes things interesting.
No. But it is an cyclic additive group for all $$n$$. And for each $$n$$ , you can consider the multiplicative group of units, $$\Bbb Z_n^×$$, which, for $$p$$ prime coincides with $$\Bbb Z_p^*$$, the multiplicative group of nonzero elements.