TITLE: Confusion about the proof that the multiplicative group of a finite field is cyclic
QUESTION [6 upvotes]: $F$  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.

REPLY [6 votes]: 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.