Necessary and sufﬁcient conditions on m and n for $\Bbb F_{p^n}$ to have a subﬁeld isomorphic with $F_{p^m}$. It is from Milne's notes. 
I think since $[\Bbb F_{p^n}:\Bbb F_{p^m}]=[\Bbb F_{p^n}:\Bbb F_p]/[\Bbb F_{p^m}:\Bbb F_p]=n/m$. If we have such a subfield, we must have $m$ divides  $n$.
But I am not able to see that if $m$ divides $n$, then we always have such a subextension.
So may I please ask how can I do it. And how to see if there is some subextensions which are isomorphic? Thanks!
 A: You are right for the necessary condition.
Assume that $m\vert n$ and let define $F\colon\mathbb{F}_{p^n}\rightarrow\mathbb{F}_{p^n}$ by:
$$F(x)=x^{p^m}.$$
Then, since $\textrm{char}(\mathbb{F}_{p^n})=p$, $F$ is a ring homomorphism. Now, let define the following subset of $\mathbb{F}_{p^n}$:
$$K:=\{x\in\mathbb{F}_{p^n}\textrm{ s.t. }F(x)=x\}.$$
Since $F$ is a ring homomorphism, $K$ is clearly a subfield of $\mathbb{F}_{p^n}$. If we are able to see that $K$ has exactly cardinality $p^m$, we are done. Indeed, there is up to isomorphism one field of cardinality $p^m$, its $\mathbb{F}_{p^m}$.
To conclude let me recall the following:

Proposition. Let $k$ be a field and $G$ be a finite subgroup of $k^\times$, then $G$ is cyclic.

Proof. Let $n$ be the cardinality of $G$ and $e$ its exponent, then $e\leqslant n$. Besides, $X^e-1$ has at most $e$ roots in $k$ thus in $G$, but by definition of the exponent all the elements of $G$ are roots. Therefore, $n\leqslant e$. Finally, $e=n$ and there is an element of order $n$ in $G$, whence the result. $\Box$
Hence, the group ${\mathbb{F}_{p^n}}^\times$ is cyclic of cardinality $p^n-1$ and since $m\vert n$, one has:
$$p^m-1\vert p^n-1.$$
Therefore, there is an element of order $p^m-1$ in $\mathbb{F}_{p^n}$, say $\alpha$ and notice that:
$$K=\{1,\alpha,\cdots,\alpha^{p^m-1}\}.$$
Whence, $K$ has indeed cardinality $p^m-1$.
