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