Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $F$ be a field, and consider the group $G=F^\times$ with multiplication ($F^\times=F-\{0\}$). How to show that for any $t\in\mathbb{Z}_{>0}$, $G$ has at most $t$ elements of order $t$?

share|improve this question

2 Answers 2

up vote 6 down vote accepted

Let $a \in F^{\times}$ be an element of order $t$. Then $a^t=1$. But the polynomial $x^t-1$ has at most $t$ roots over $F$.

share|improve this answer

So here $F$ is a finite field. We know that the group of units in a finite field is cyclic, and thus if $|F|=q=p^n$ then the group of units is a cyclic group of order $q-1$.

Also note that by induction we can show that over any field a polynomial of degree $d$ has at most $d$ roots.

Using both of these, you should try to answer your question.

share|improve this answer
    
The claim is true for any field, not necessarily finite. –  DonAntonio Dec 3 '12 at 2:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.