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Let p be a prime and let $\mathbb{Z^*{_p}}$ = $\{1,2,...p-1\}$. Show that $\mathbb{Z^*{_p}}$ is group closed under multiplication.

$\mathbb{Z^*{_p}}$ is associative because $\mathbb{Z}$ is associative under multiplication.

The identity is just 1.

The inverse: let m$ \in \mathbb{Z^*{_p}}$ Need to find $m' \in \mathbb{Z^*{_p}}$ such that $m*m'$ = 1 mod p$

$m*m' = 1 mod_p \iff m*m'+p*r = 1$ , $r\in \mathbb{Z}$

Now for each $m \in \mathbb{Z^*{_p}}$, gcd(m,p) = 1 and I'm supposed to use this somehow. But this is where I need help.

Mathaniel
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  • Recall that $\bar{n}\in \mathbb{Z}_p$ is invertible if and only if $\text{gcd}(n,p)=1$, indeed this follows from Bezout's theorem. If $p$ is a prime, then $\text{gcd}(n,p)=1$ for all $1\leq n \leq p-1$. It follows that $\bar{n}$ is invertible, hence $\bar{n}\in \mathbb{Z}_p^*.$ – Mathematician 42 Mar 14 '17 at 08:23
  • I'm a bit confused by what you mean by showing that $\mathbb{Z}_p^$ is a subgroup. Of what should it be a subgroup? Also, by definition there are multiplicative inverses, that's the meaning of the $$. – Mathematician 42 Mar 14 '17 at 08:24
  • i meant group. fixed now. – Mathaniel Mar 14 '17 at 08:27

1 Answers1

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Let $\bar{m}\in \mathbb{Z}_p^*$. Then $\text{gcd}(m,p)=1$. By Bezout's theorem there exist numbers $\alpha, \beta\in \mathbb{Z}$ such that $\alpha m+\beta p=1$. Looking at the last equation module $p$, we get that $\overline{\alpha m}=\overline{1}$. Hence $\overline{\alpha}=(\overline{m})^{-1}$. Thus inverses exist.

Mathematician 42
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