A Finite Field of Order $125$ has a Subfield of order $25$?

Question

How to prove that every finite field of order $125$ has a subfield of order $25$. In general what is the strategy to attack such kind of problems?

Answer 1

Consider a field $F$ with $q$ elements, $q$ being a power of some prime. Suppose $L$ is a field containing $F$, with $[L\colon F]=m$. Since $L$ is an $F$-vector space of dimension $m$, $|L|=q^m$. Thus $[\mathbb F_{125}\colon\mathbb F_5]=3$. Now do you see why there’s no field strictly between these two fields?

Answer 2

So if there is a subfield G of size 25, you can remove the zero element and it becomes a group under multiplication. Specifically, it is a subgroup of F-{0}, but F-{0}=124, and 24 doesn't divide 124. All subgroup sizes must divide the size of the group they come from, therefore you can never have a subfield of size 25.