# Why does a multiplicative subgroup of a field have to be cyclic?

The book "A First Course in Abstract Algebra" by Fraleigh says

If $G$ is a finite subgroup of the multiplicative group $\langle F^*,\cdot\rangle$ of a field $F$, then $G$ is cyclic. In particular, the multiplicative group of all nonzero elements of a finite field is cyclic.

I wonder why that is. Take $a,b\in F$ such that $a^2=b^2=1$. Then $\{1,a,b,ab\}$ is a non-cyclic subgroup of $F$. Where am I going wrong?

Thanks in advance!

• How are you concluding that $\{1,a,b,ab\}$ is non-cyclic? What if $a=b$? What if $F=\mathbb{R}$, $a = 1$, $b = -1$? Then $\{1,a,b,ab\} = \{1,-1\}$ is certainly cyclic. – Amit Kumar Gupta Jan 27 '14 at 5:08
• The quadratic $\,x^2-1\,$ cannot have $\,4\,$ distinct roots $\{1,a,b,ab\}$ in a field. – Bill Dubuque Jan 27 '14 at 5:09
• Your error is in presuming each element of $\{1,a,b,ab\}$ is distinct from the other three. – anon Jan 27 '14 at 5:10
• My initial assumption (which I realised I haven't mentioned) is that $a\neq b$. – fierydemon Jan 27 '14 at 5:13
• Ayush, you just asked why $1$ has to be a root of $x^2-1$. Surely that's obvious... – anon Jan 27 '14 at 5:27

## 4 Answers

The quadratic $\,x^2-1\,$ cannot have $\,4\,$ distinct roots $\{1,a,b,ab\}$ in a field. This property of fields plays a key role in the proof - see below.

Theorem $\$ A finite subgroup $\rm\:G\:$ of the multiplicative group of a field is cyclic.

Proof $\$ The proposition below yields, with $\rm\,m = maxord(G) = expt(G),\,$ that $\rm\, x^m = 1\,$ has $\rm\:\#G\:$ roots. Since a polynomial $\rm\:f\:$ over a field satisfies $\rm\:\#roots\ f \le deg\ f\:$ we infer that $\rm\: \#G \le m.\:$ But maxorder $\rm\:m \le \#G\:$ since $\rm\:g^{\#G} = 1\:$ for all $\rm\:g \in G\:$ (Lagrange). $\:$ Thus $\rm\:m = \#G = maxord(G),\:$ therefore $\rm\:G\:$ has an element of order $\rm\#G,\:$ hence $\rm\:G\:$ is cyclic.

$\begin{eqnarray}\rm{\bf Proposition}\quad maxord(G) \!&\,=\,&\rm expt(G)\ \text{ for a finite abelian group}\ G,\ i.e.\\ \\ \rm max\ \{ ord(g) : \: g \in G\} \!&\,=\,&\rm min\ \{ n>0 : \: g^n = 1\ \ \forall\ g \in G\}\end{eqnarray}$

Proof $\$ By the lemma below, $\rm\: S\, =\, \{ ord(g) : \:g \in G \}$ is a finite set of naturals closed under$\rm\ lcm$.

Hence every $\rm\ s \in S\:$ is a divisor of the max elt $\rm m\$ [else $\rm\: lcm(s,m) > m\,$],$\$ so $\rm\ m = expt(G)$.

Lemma $\$ A finite abelian group $\rm\:G\:$ has an lcm-closed order set, i.e. with $\rm\: o(X) =$ order of $\rm\: X$

$$\rm X,Y \in G\ \Rightarrow\ \exists\ Z \in G:\ o(Z) = lcm(o(X),o(Y))$$

Proof $\ \$ By induction on $\rm\: o(X)\, o(Y).\$ If it's $\:1\:$ then trivially $\rm\:Z = 1$. $\$ Otherwise

write $\rm\ o(X) =\: AP,\: \ o(Y) = BP',\ \ P'|\,P = p^m > 1,\$ prime $\rm\: p\:$ coprime to $\rm\: A,B.$

Then $\rm\: o(X^P) = A,\ o(Y^{P'}) = B.\$ By induction there's a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$

so $\rm\ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP')\: =\: lcm(o(X),o(Y)).$

• If you leave some details for the OP to work out on his own then you get essentially the same proof as in my answer. :) – Amit Kumar Gupta Jan 27 '14 at 5:33
• @Amit The point of my comment was to help readers understand the relationship between the answers - nothing more. Otherwise, in threads with many related answers, a student might expend much effort before realizing that many answers are essentially duplicates. Here a reader can decide on a sketch or a complete proof, whatever works best. – Bill Dubuque Jan 27 '14 at 5:55
• Yes I know, just being snarky. – Amit Kumar Gupta Jan 27 '14 at 5:57

Let $G < \langle F^\ast, \cdot\rangle$ be a finite subgroup. It suffices to show that $\mathrm{ord}(x) = |G|$ for some $x \in G$, since such an $x$ can be taken to be a generator for $G$.

Suppose not, and that the maximum order of any element of $G$ is some $n < |G|$. It's easy to see that $x^n = 1$ for all $x \in G$, indeed, $n$ must be the least common multiple of the orders of the elements of $G$ (this fact doesn't rely on the assumption $n < |G|$). Thus every element of $G$ is a solution to $x^n = 1$, of which there are at most $n$ in the field $F$. So

$$\left|G\right| \leq \left|\{x\in F : x^n = 1\}\right| \leq n < \left|G\right|$$

contradiction.

• If you complete the omitted parts then you get essentially the same proof as in my answer. – Bill Dubuque Jan 27 '14 at 5:27
• Finally a clear answer : ) – Psylex Jun 1 at 20:41

Sketch: $F$ a field implies $F[x]$ a Euclidean domain implies $F[x]$ a UFD. Since $F$ is a field, all the $F$ are the units in $F[x]$, so no degree zero polynomial is an irreducible in $F[x]$. It is an exercise to show degree 1 irreducible polynomials have a single root in $F$. It is an exercise to show that degree $\geq$2 irreducible polynomials have no roots in $F$.

Consider a polynomial of degree $d$ in $F[x]$. This polynomial is the product of a unit and powers of irreducibles in $F[x]$. The sum of the polynomial degrees of the unit (i.e. zero) and the irreducibles (counted with multiplicity) is the degree of the polynomial because $F[x]$ is a Euclidean domain. Therefore, the polynomial has $\leq d$ roots in $F$.

Let $n = |G|$ and $m = \gcd(\{|g| \, \mid \, g \in G\})$. By Lagrange, every element's order divides $n$. The least number divided by every element's order is $m$, so $m \leq n$. Consider now the polynomial $P(x) = x^m-1$, which has degree $m$ and therefore has $\leq m$ roots in $F$. Since every element of $G$ satisfies the equation $P(x)=0$, $n \leq m$. Therefore $n = m$.

(One variant of this argument uses Euler's totient here. I think this is usually attributed to Serre. Can anyone confirm/deny that?)

Let $p$ range over the primes dividing $n$ such that $p^k$ exactly divides $n$ and let $\mathrm{Syl}_p$ be a Sylow $p$-subgroup of $G$. If $\mathrm{Syl}_p$ is not cyclic, then every element of $\mathrm{Syl}_p$ has order $\leq p^{k-1}$. But then the maximum power of $p$ in the factorization of $m$ is $p^{k-1}$ and that in $n$ is $p^k$. Since $m=n$, $\mathrm{Syl}_p$ is cyclic. For distinct primes $p$, $q$, let $g_p, g_q$ generate $\mathrm{Syl}_p$,$\mathrm{Syl}_q$, respectively. We find that $(|g_p|,|g_q|)=1$ so $g_p g_q$ generates a cyclic group with order the product of the orders of $\mathrm{Syl}_p$ and $\mathrm{Syl}_q$. Since $F$ is a field, $G$ is abelian so $G$ is the direct product of its Sylow subgroups. By a quick induction, the product of a set of generators, one each from the Sylow $p$-subgroups as $p$ ranges over primes dividing $n$, generates $G$.

• I vaguely recall seeing the totient version credited to someone besides Serre, but I cannot recall who at the moment. – Bill Dubuque Jan 27 '14 at 6:06
• That $\gcd$ expression was a little surprising, I think because I confused the \mid in the set comprehension with the bars on the order-of-$g$ notation. But I wasn't sure what I would have preferred. Then I wondered about $$\gcd_{g\in G}(\lvert g\rvert).$$ – MJD Jan 27 '14 at 6:19

Hint A finite abelian group is cyclic iff its order equals its exponent.