Show that $\mathbb{Z}_p\setminus\{\overline{0}\}$ is not a group if $p$ is not prime.

Question

The answer is too short that I think I've gone wrong at some point!

Q: If $p$ is prime, then the nonzero elements of $\mathbb{Z}_p$ form a group of order $p-1$ under multiplication. Show that this statement is false if $p$ is not prime.

A: $p$ is not prime, so take it $p=4$. Then $\mathbb{Z}_4 \backslash \{\bar{0}\}=\{\bar{1},\bar{2},\bar{3}\}$. It is a monoid, but since $\bar{2}$ does not have an inverse element, then $\mathbb{Z}_4 \backslash \{\bar{0}\}$ is not a group.

Answer 1

Hint: if $n$ is composite, then the set of non-zero elements of $\mathbb{Z}_n$ is not closed under multiplication.

Answer 2

Let $n>1$ be a non-prime natural number. Therefore there exist $a,b \in \{2, 3 ,\ldots n-1\}$ such that $p = ab$ and consequently $0 = ab$ in ${\mathbb Z}_n$. Then one can see that $a,b$ do not have inverse, since otherwise, if $a'$ and $b'$ be their inverse, one have that $a.a'=b.b'=1$ from one hand, and from other hand $ab=0$, where these together means that $a=b=0$ which contradicts $a,b \in \{2, 3 ,\ldots n-1\}$. So ${\mathbb Z}_n$ can not be a multiplicative group.