I'm trying to prove that $\mathbb Z_p^*$ ($p$ prime) is a group using the Fermat's little theorem to show that every element is invertible.

Thus using the Fermat's little theorem, for each $a\in Z_p^*$, we have $a^{p-1}\equiv1$ (mod p). The problem is to prove that p-1 is the least positive integer which $a^{p-1}\equiv1$ (mod p).

Remark: $\mathbb Z_p^*$ is $\{\overline 1,...,\overline {p-1}\}$ with multiplication.

I need help.

Thanks a lot.

  • 3
    $\begingroup$ Why should it be the least positive integer? You have just proved that $\overline{x}^{-1}=\overline{x}^{p-2}$, which gives the inverse. $\endgroup$ – egreg Apr 20 '13 at 9:59
  • $\begingroup$ As far as I know, this notation is for the group of units of the ring $\mathbb{Z}_p$. So it is a group. But it turns out that $\mathbb{Z}_p^*=\{a\in\mathbb{Z}_p\;;\; a\neq 0\}$. That is: $\mathbb{Z}_p$ is a field. And that is what you are trying to prove: every nonzero element is invertible. $\endgroup$ – Julien Apr 20 '13 at 10:00
  • $\begingroup$ @egreg yes, you're right, I didn't think in this way. Thank you very much $\endgroup$ – user42912 Apr 20 '13 at 10:04
  • $\begingroup$ @julien yes, I know, thank you $\endgroup$ – user42912 Apr 20 '13 at 10:04

You can't show that $p-1$ is the least positive integer $r$ such that $a^r\equiv 1\pmod{p}$, because in general it isn't: for instance, the least integer for $a=1$ is $1$.

But all you need is to find an element which acts as an inverse:

$$a\cdot a^{p-2} \equiv 1 \pmod{p}$$

so that, for any $\overline{x}\in\mathbb{Z}^*_p$ you have

$$\overline{x}\cdot\overline{x}^{\,p-2} = \overline{1}$$

and so


  • $\begingroup$ I have a question. How can we know that $\overline{x}^{\,-1}=\overline{x}^{\,p-2} \in \mathbb{Z}_{p}^{*}$? Thanks for your answer. $\endgroup$ – OpiRabbit Sep 14 '13 at 21:21
  • $\begingroup$ @nameless $\mathbb{Z}^*_p$ is a group (with respect to multiplication) and has $p-1$ elements. If a group $G$ has $n$ elements, then $g^n=1$ for any $g\in G$ and therefore $g^{-1}=g^{n-1}$. $\endgroup$ – egreg Sep 14 '13 at 21:25
  • $\begingroup$ Yes. Now, we want to show $(\mathbb{Z}_{p}^{*}, \cdot)$ is a group. For show it has an inverse, we find $\overline{x}^{p-2}$ such that $\overline{x}\cdot\overline{x}^{\,p-2} = \overline{1}$. But we need to show that $\overline{x}^{p-2}$ lies in $\mathbb{Z}_{p}^{*}$ ? $\endgroup$ – OpiRabbit Sep 14 '13 at 21:30

Please check my inverse solution.

Let $\overline{x} \in \mathbb{Z}_{p}^*$. Then $(x,p)=1$, that is there exist $k, q \in \mathbb{Z}$ such that $kx + qp=1$. This implies $\overline{k}\overline{x}=\overline{1}$. Suppose that $\overline{k}=\overline{0}$. Then $\overline{0}=\overline{1}$, a contradict. Thus $\overline{k} \not = \overline{0}$, so $k \not = 0$.

Now, we write $k=mp+r$ where $0 \leq r < p$. If $k<p$, then $(k,p)=1$. It follows that $\overline{k} \in \mathbb{Z}_{p}^*$. If $k=p$, then $r=0$. It follows that $\overline{k}=\overline{0}$, a contradict, thus $k \not = p$. If $k>p$, then we have $k=mp+r$ where $0 < r < p$. Hence $\overline{r} \in \mathbb{Z}_p^*$. Since $k \equiv r \, (\mod p)$, $\overline{k} \in \mathbb{Z}_p^*$.

Therefore $\mathbb{Z}_p^*$ has inverse.


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