$F$ 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.
Best Answer
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.