Question

Let p be prime and let b be a nonzero element of ${\mathbf{\mathbb{Z}}}_{\mathbf{p}}$. Show that ${\mathit{b}}^{\mathbf{p}\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{1}$[Hint Theorem 7.16]

Short answer

It is proved that$b^{p-1}=1$ .

Step-by-step solution

Revisit Theorem 7.16

Let F be any one of $\mathrm{\mathbb{Q}},\mathrm{ℝ},\mathrm{ℂ},{\mathrm{\mathbb{Z}}}_p$( with p prime), and let $F^{*}$be the multiplicative group of non-zero element. If G is a finite subgroup of $F^{*}$, then G is cyclic.

Prove that bp-1=1

Since p is prime and from the theorem, we can say that ${{\mathrm{\mathbb{Z}}}_p}^{*}$is cyclic of order $p-1$.

Now, let us suppose a is an element of${{\mathrm{\mathbb{Z}}}_p}^{*}$ , $a\in {{\mathrm{\mathbb{Z}}}_p}^{*}$be a generator.

For any$b\in {{\mathrm{\mathbb{Z}}}_p}^{*}$ , there must be some order n, so$0\le n<p-1$ such that$b=a^n$

Since the order of ${{\mathrm{\mathbb{Z}}}_p}^{*}$is $p-1$, $b^{p-1}$becomes:

$$\begin{gathered} b^{p-1}={\left(a^n\right)}^{p-1} \\ ={\left(a^{p-1}\right)}^n \\ =1^n \\ =1 \end{gathered}$$

Hence, it is proved that$b^{p-1}=1$