For each divisor $m$ of $n$, $GF(p^n)$ has a unique sub field of order $p^m$ .
The proof of this theorem in Gallian goes like this :
Suppose that $m$ divides $n$. Then, since :
$(p^n-1) = (p^m-1)(p^{n-m})+p^{n-2m})+\cdots+p^{m}+1)$
Hence : $p^m-1$ divides $p^n-1 \implies p^n-1=(p^m-1)t$ for some $t$.
Let $K=\{ x \in GF(p^n)~~|~~x^{p^m}=x\}$
it can be now shown that $K$ is a sub field of $GF(p^n)$, we have $|K| \leq p^m$.
Let $\langle a \rangle=GF(p^n)^*$. Then $|a^t| = p^m-1$
Since, $(a^t) p^{m-1} =1 \implies a^t \in K$
Hence, $K$ is a sub field of $GF(p^n)$ of order $p^m$.
I am not able to understand how $(a^t) p^{m-1} =1$. And how does that ensure that $a^t \in K$? Please give me some hints.
Thank you for your help..