Multiplicative $\mathbb{Z}^{*}_p$

Question

Question: How to prove that $\mathbb{Z}^{*}_p$ is a group under multiplication operation where $p$ is prime.

Proof: We define $$\mathbb{Z}^{*}_p=\{[a] \in \mathbb{Z}_p: [a] \neq [0]\}$$

1. For well-defined: By assumption $[a] \neq [0]$ so $p| a$ thus $(a, p)=1$.

Let $[a], [b] \in \mathbb{Z}^{*}_p$. Then $$[a][b]=[ab]=[0]$$

$$ab \equiv 0\pmod{p}$$

so, $$ p|ab$$ since $(a, p)=1$ and $p|a$. This contradicts to the assumption that $[b] \neq [0]$ thus $[ab] \neq [0]$ therefore $[ab] \in \mathbb{Z}^{*}_p$

2. Associative law Let $[a], [b], [c] \in \mathbb{Z}^{*}_p$. Then $$[a]([b][c])=([a][b])[c]$$

3. Identity Element

4. Inverse Element

How to check associativity, identity and inverse so that $\mathbb{Z}^{*}_p$ forms a group under multiplication operation.

Answer 1

Your attempt for (1) is almost correct in the math part, but is poorly written. I will rewrite what you did in a rigorous way:

By contradiction, let us assume that there exists $[a],[b]\in\mathbb{Z}^*_p$, such that $[a][b]\notin\mathbb{Z}^*_p$. Then $[a][b]=[ab]=[0]$, which implies $ab\equiv 0$ (mod p), so $p\mid ab$. Since $p$ is a prime number, either $p\mid a$ or $p\mid b$, which is equivalent to say that $[a]=[0]$ or $[b]=[0]$. Hence, $[a]\notin\mathbb{Z}^*_p$ or $[b]\notin\mathbb{Z}^*_p$, contradiction.

For (2), (3) and (4), the only difficult part is the existence of an inverse (since associative and identity follows from the associative and identity of $\mathbb{Z}$), to do this let $[n]\in\mathbb{Z}_p^*$, WLOG, we can say that $n\in\lbrace 1,\dots,p-1\rbrace$. Since $p$ is prime $\gcd(n,p)=1$. Applying Bezout's identity there is some $a,b\in\mathbb{Z}$ such that $an+bp=\gcd(n,p)=1$. From this we have immediately that $[a]$ is the inverse element of $[n]$.