TITLE: Let F be a field of order 32. Show that the only subfields of F are F itself and {0,1}.
QUESTION [6 upvotes]: $F$ is a field of order $32$.
$F$ and {$0,1$} are trivial subfields of $F$.
But how can we show that these are the only subfields of $F$?
Can someone give me a direction to this question?

REPLY [3 votes]: If there is a field $F$ of order $p^n$, $p$ is a prime & $n$ is a positive integer, then for each divisor $m$ of $n$ there is a unique subfield of order $p^m$. These are the only subfields of $F$.
Now, we have $32=2^5$. The divisors of $5$ are $1,5$. So there are only two subfields, one is of order $2^1$ and another is of order $2^5$. And hence the only subfields are {0,1} and the field itself.