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How to show if $p$ is prime, then group $Z_{p}^{*}$ is cyclic.

Tip

  1. Let $g$ and $h$ of a commutative group $G$ have orders $n$ and $m$ respectively. There exists and element $x \in G$ of order $LCM (n,m)$
  2. In any field $\mathbb{K}$ a polynomial $f \in \mathbb{K} [x]$ of degree $n$ has at most $n$ different roots.

I have no idea.

new_student
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  • helpful site : http://math.stackexchange.com/questions/290427/is-mathbb-z-p-1-2-3-p-1-a-cyclic-group – HK Lee Apr 18 '15 at 13:16

2 Answers2

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(In the given links I did not find a proof using exactly your hints that's why I provided this proof, but you can see other methods too)

Proving that $\Bbb Z_p^*$ is cyclic is equivalent to proving the existence of an element of order $p-1$, take $l=\text{lcm}(\text{ord}(1),\text{ord}(2),\cdots,\text{ord}(p-1))$ so it's clear that: $$\forall k\in \Bbb Z_p^* \ \ \ \ \ \ \ k^l=1 $$ because $\text{ord}(k)|l$ for every $k$, This means that $x^l-1$ is a polynomial of degree $l$ in the field $\Bbb Z_p$ which has more than $p-1$ roots, as a consequence $l\geq p-1$.

Now because there exists an element of order $LCM(\text{ord(a)},\text{ord}(b))$ for every two elements (This is the first hint: to prove it take just the product $ab$) there exists an element $g$ of order $l$ hence $l\leq p-1$.

Finally there exits an element $g$ of order $l=p-1$, thus $\Bbb Z_p^*$ is cyclic.

modnar
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Elaqqad
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  • This is a special case of a proof that a finite subgroup of the multiplicative group of a field is cyclic, e.g. see here. – Bill Dubuque Apr 18 '15 at 14:18
  • I'm sorry I did not see your answer – Elaqqad Apr 18 '15 at 14:58
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    There will always be some amount of overlap or duplication on a site this large. No need to apologize for that, esp, considering that special cases often prove instructive. My comment was merely meant to inform readers about the generalization. – Bill Dubuque Apr 18 '15 at 15:02
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We have the much more general

Proposition. Any finite subgroup of the multiplicative group of a field is cyclic.

Proof. Let $F$ be a field and $G$ a finite subgroup of $F^\times$. Let $n=|G|$. For $m\in\mathbb N$, let $f(m)$ denote the number of distinct $m$th roots of unity in $F$ and let $g(m)$ denote the number of distinct primitive $m$th roots (i.e., those that are not $d$th roots for any $d<m$). Then clearly $$\tag1 f(m)=\sum_{d\mid m}g(d).$$ By Möbius inversion it follows that $$\tag2 g(m)=\sum_{d\mid m}\mu(m/d)f(d).$$ Since $m$th roots of unity are roots of $P_m(X):=X^m-1$, we have $f(m)\le m$ for all $m$. If $d\mid m$ then $$P_m(X)=P_d(X)(1+X^d+X^{2d}+\ldots+X^{m-d})$$ shows that if $P_m$ has $m$ distinct roots, the same must hoild for $P_d$, i.e., if $f(m)=m$ then also $f(d)=d$. Since $a^n=1$ for all $a\in G$, we conclude $f(n)=n$ and hence $f(d)=d$ for all $d\mid n$. Using $(2)$, we obtain $$ g(n)=\sum_{d\mid n}\mu(n/d)d=\phi(n)>0.$$ Thus $G$ contains at least one element $a$ of order $n$ and hence $G=\langle a\rangle$ is cyclic. $_\square$

Hagen von Eitzen
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