# A field $K$ of order $q=p^r$ contains a subfield $K'$ of order $q'=p^k$ if and only if $k\mid r$.

I was trying to prove the following theorem:

A field $$K$$ of order $$q=p^r$$ contains a subfield $$K'$$ of order $$q'=p^k$$ if and only if $$k\mid r$$.

My attempt at the proof was:

$$K'\setminus\{0\}$$ is a subgroup of $$K\setminus\{0\}$$. So, $$p^k-1\mid p^r-1$$.

From the exponent gcd lemma [$$\gcd(a^x-1,a^y-1)=a^{\gcd(x,y)}-1$$ when $$a,x,y\in\mathbb N,a\geq 2$$.]

We have that $$k\mid r$$.

$$K'\setminus\{0\}$$ is a subgroup of $$K\setminus\{0\}$$.
• Then you still need to prove that if $k \vert r$ then an appropriate subfield exists. – Robert Shore May 15 at 9:06