$\mathbb{Z}^*_p$ is a group under multiplication

Question

Show that the set $$ \mathbb{Z}^*_p=\{1,2, \dots, p-1\} $$ where $p \in \mathbb{N}$ is prime, is a group under multiplication.

Attempt:

Associativity and identity:

Obviously, multiplication $\bmod p$ is associative and $1 \in \mathbb{Z}^*_p$ (identity element).

Inverses:

Let $k \in \mathbb{Z}^*_p$. Since $k<p$, $\gcd(k,p)=1$ and therefore Bezout's Lemma ensures that there exist $a,b \in \mathbb{N}$ s.t. $$ ak+bp=1\iff ak+bp \equiv 1 \space\bmod p \iff ak \equiv1 \bmod p $$

However, is the fact $a<p$ guaranteed, so that $a=k^{-1}\in \mathbb{Z}^*_p?$

Is it because the congruence class $\bar{a}$ with $a<p$ contains every $a_i$ you choose s.t. $ a_ik \equiv 1 \bmod p $?

Answer 1

I think you are mixing up integers and classes of integers modulo $p$. The set you just gave should not be set of integers, but a set of classes of integers modulo $p$. Indeed, it if was a set of integers, it would not be a group : $(p-1) \times 2$ is not in this set !

But if you consider these elements as classes of integers modulo $p$, then all of your reasoning makes sense, and the element $a$ you get is indeed the inverse you are looking for : it's a class of integers. It just turns out that there is a representative of it between $1$ and $p-1$.