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Suppose $K$ is a finite field and $F_1$, $F_2$ are subfields of $K$ having the same number of elements. Then $F_1 = F_2$.

I have seen several posts on MSE about this fact (e.g., here and here), and their answers all use the polynomial $X^{p^n}-X$ to prove the fact. But it seems to me that one could prove the fact by considering merely the multiplicative group of these fields: since $K^*$ is cyclic, it contains only one subgroup of each order $d \mid |K^*|$, hence $F^*_1$ and $F^*_2$ coincide, so are $F_1$ and $F_2$.

Is there anything wrong with the latter proof? Why does the standard proof refer to $X^{p^n}-X$?

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    $\begingroup$ Looks like it's correct. However, the statement is true even without the assumption that $K$ is finite. If $K$ is any field, then for any natural number $q$, $K$ contains at most one subfield with $q$ elements. If $K$ is not finite, its multiplicative group is not cyclic, so the same argument will not work. But the proof with the polynomial $x^{p^n}-x$ still works without change. $\endgroup$ Commented Jul 23, 2023 at 8:56
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    $\begingroup$ @Mark it could still be made to work by using that any finite subgroup of the multiplicative group of a field is cyclic $\endgroup$ Commented Jul 23, 2023 at 9:30
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    $\begingroup$ @doetoe But how does it help? A group can contain many different cyclic subgroups of the same order. It's just that a group which is itself finite cyclic contains at most one cyclic subgroup of each finite order, which is what OP used. $\endgroup$ Commented Jul 23, 2023 at 9:42
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    $\begingroup$ @Mark $F_1^\ast$ and $F_2^\ast$ generate a finite subgroup $H$ of $K^\ast$, and $H$ is therefore cyclic itself. Since $F_1^\ast$ and $F_2^\ast$ are subgroups of it of the same order, they have to be equal. $\endgroup$ Commented Jul 24, 2023 at 10:23

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