6

$F$ is a finite field and $F^{*}$ is the multiplicative group of $F$. Then $F^{*}$ is cyclic.

The method that they use here to prove it is that if for each $d$ such that $d| |F^{*}|$ we can show $F^{*}$ has only one cyclic subgroup of order $d$, then $F^{*}$ is cyclic.

Let $d$ be a divisor of $|F^{*}|$ and $C$ is the subgroup of order $d$. then every element of $C$ satisfies the equation $x^{d} =1$ which can have at most $d$ solutions so $C$ is the only group of order $d$ as if there were any other subgroup , say , $H$ of order $d$ then that will have elements different from that of $C$ which also will satisfy the equation $x^{d}=1$ and thus implying that $x^{d}=1$ has more than $d$ solutions which is not possible.

Now here is my problem : there are many groups that have more than $1$ cyclic subgroups of the same order and all of those elements of those groups satisfy the same equation . How does that work there and why can the same thing be applied here as a contradiction $?$

I hope I have conveyed my problem clearly.

Thanks for any help.

hardmath
  • 37,715
  • 1
    Isn't the word finite in front of field pretty important here? After all the multiplicative group of nonzero rationals is not cyclic, etc. – hardmath Sep 13 '15 at 20:36
  • @hardmath: Not really. The word finite in front of subgroup of the multiplicative group OTOH is crucial. The multiplicative group of nonzero rationals has $\langle -1\rangle$ as its ONLY finite subgroup. Inside $\Bbb{C}$ we have finite cyclic subgroups of any order, consisting of roots of unity. In a finite field, of course, all the nonzero elements are roots of unity. – Jyrki Lahtonen Sep 13 '15 at 20:49
  • 1
    @JyrkiLahtonen: I was referring to the omission of finite in the title and first paragraph, which claim "the multiplicative group of a field is cyclic". Thus the counterexample of field $\mathbb{Q}$, whose multiplicative group is not cyclic. – hardmath Sep 14 '15 at 02:43
  • @hardmath: Ok. Sorry. Apparently I didn't read the question carefully enough. – Jyrki Lahtonen Sep 14 '15 at 05:17
  • 1
    @JyrkiLahtonen: I'm going to make the edit for the sake of clarity, since the OP implicitly assumes $F^$ is finite in the second paragraph. But your comment got me to wondering about the converse, if $F^$ is cyclic, then field $F$ is finite, which I plan to post as a new Question. – hardmath Sep 15 '15 at 01:29
  • If $F^$ ha a cyclic subgroup with $d$ members whenever $d | |F^| $, this includes the case $d=|F^*|$. – DanielWainfleet Sep 15 '15 at 02:09
  • 1
    @hardmath: That has been asked many times, so ... Oh, you already did :-( – Jyrki Lahtonen Sep 15 '15 at 04:45

1 Answers1

6

The key here is that the finite group is a subgroup of a multiplicative group of a field. In a field $F$ the polynomial $p(x)=x^d-1$ can have at most $d$ zeros. This is because the ring $F[x]$ is a UFD, and each zero $\alpha$ of $p(x)$ gives rise to a linear factor $x-\alpha\mid p(x)$.

OTOH, in a multiplicative group it is perfectly possible for the equation $x^d=1$ to have more than $d$ solutions.

Note that it is not essential for the field to be finite itself as long as the group is finite. The same holds for all fields: A finite subgroup of the multiplicative group of any field is cyclic.

Jyrki Lahtonen
  • 140,891