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!  
 A: $\,x^2-1\,$ cannot have $\,4\,$ distinct roots $\{1,a,b,ab\}$ since a quadratic has at most two roots in a field (or domain). This property 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 $\rm\,f\neq 0\,$ 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  $\rm\ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP')\: =\: lcm(o(X),o(Y)).$
A: 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.
A: 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 $G$.
A: Hint A finite abelian group is cyclic iff its order equals its exponent. 
