# Proving that a set is a group and that it's closed under multiplication(basic group theory)

let p be a prime and let $\mathbb{Z}_p^* = \{1,2...p-1 \}$ Supposed to show that $\mathbb{Z}_p^*$ is a group and that it's closed under multiplication.

Associativity: since $\mathbb{Z}_p^* \leq \mathbb{N}$ and $\mathbb{N}$ is associative, $\mathbb{Z}_p^*$ is associative by inheritence.

Identity: try $1: 1*a = a = a*1$ for all $a \in \mathbb{Z}_p^*$

Inverse: Let $a\in \mathbb{Z}_p^*$, must find $a^{-1}a =1$ for $a^{-1} \in \mathbb{Z}_p^*$

$a^{-1}a = 1mod_p \iff a^{-1}a + pr =1$ , for some $r\in \mathbb{Z}$

since and p are co primes: $gcd(a,p) =1$ and $\alpha a * \beta p =1$ for $\alpha , \beta \in \mathbb{Z}$ by Bezout's lemma.

plugging in we get $\alpha a = 1 mod_p \iff \alpha a = 1$

$a=\alpha^{-1}$ with $\alpha = a^{-1}$

$a= (a^{-1})^{-1} = a$ (Is this sufficient/enough?)

Closure: let $a_1,a_2 \in\mathbb{Z}_p^*$, must show $a_1a_2 \neq 0 mod_p$

Assume for contradiction: $a_1a_2 = 0 mod_p$

$a_1a_2 = rp$ for some r $\in \mathbb{Z}$

$\iff a_1a_2 = rp \iff r= \frac{a_1a_2}{p}$ but both $a_1$ and $a_2$ are co-primes to p by definition and so the equation has no solution for r and hence $\mathbb{Z}_p^*$ is closed under multiplication.

This means that $\mathbb{Z}_p^*$ is a group under multiplication. Have I done anything wrong?



1. "since $\mathbb{Z}_p^* \leq \mathbb{N}$". No, this is not a subgroup. First of all, $\mathbb{N}$ is not a group. Secondly, you need not show associativity, because you said "Supposed to show that $\mathbb{Z}_p^*$ is closed under multiplication." This means something else.
2. The "definition" $\mathbb{Z}_p^* = \{1,2...p-1 \}$ is not complete. What is the group operation? Multiplication? Then $(p-1)(p-1)>p-1$ for $p>2$ is not in the set. So it is not closed. Same with addition.
• Associativity follows by the definition of the multiplication of the equivalence classes $\overline{x}$ in $\mathbb{Z}_p^{\ast}$. – Dietrich Burde Mar 17 '17 at 11:42
• Well, I think the idea is right, but for better answers see the duplicate. First you should correct your definition of $\mathbb{Z}_p^{\ast}$ by using equivalence classes. Actually, it is the group of units of the quotient ring $\mathbb{Z}/p\mathbb{Z}$. – Dietrich Burde Mar 17 '17 at 12:06