The multiplicative group of nonzero elements of a finite field is cyclic (contrapositive proof)

Question

I have seen a few proofs of this claim, but I am interested in the proof by contrapositive. How can I show that if the multiplicative group is not cyclic then for some $n>0$, $x^n=1$ has more than n solutions in F, the finite field? Then I can show that since $x^n=1$ has at most n solutions (or is it exactly n solutions?) in F, the multiplicative group is cyclic.

See e.g. this method of proof in the linked dupe. – Bill Dubuque Nov 15 '22 at 11:07 — Bill Dubuque, Nov 15 '22 at 11:07

score 3 · Answer 1 · edited Nov 15 '22 at 12:05

3

$F^\times$ has $k$ elements. If it is not cyclic then (since it is abelian) the lcm $m$ of the order of its elements is smaller than $k$. So $x^m-1$ has $k>m$ roots in $F$.

edited Nov 15 '22 at 12:05

Anne Bauval

34,650

answered Nov 15 '22 at 09:54

reuns

77,999

1

@AnneBauval Obviously $F^\times$ is abelian – reuns Nov 15 '22 at 11:51
Oh yes, "obviously ;-)" (I forgot that the french translation of "field" is "corps commutatif"). – Anne Bauval Nov 15 '22 at 12:03

The multiplicative group of nonzero elements of a finite field is cyclic (contrapositive proof)

1 Answers1