Could $E$ have a subfield of order 8

Question

Consider the irreducible polynomial $g = X^4 + X + 1$ over $F_2$ and let $E$ be the extension of $F_2 =$ {0, 1} with a root $α$ of $g.$

Could $E$ have a subfield of order 8

When working through my work Ive come across this question which has stumped me and I cant get my head aound if this is true of not so any help will be appreciated.

Answer 1

If $F$ is a subfield of $E$ and both fields are finite, we always have $|E|=|F|^n$ where $n$ is the degree of the field extension. The reason is that $E$ has structure as an $F$-vector space.

In particular a proper subfield of order $8$ must live in a field with at least $64$ elements...

Answer 2

$E$ has $2^4=16$ elements and so the multiplicative group $E^\times$ has order $15$.

A subfield of order $8$ would yield a multiplicative group of order $7$, which does not divide $15$, as required by Lagrange's theorem.

Therefore, there is no subfield of size $8$.

This argument implies that the only possible sizes for proper subfields of $E$ are $2$ and $4$, since the proper divisors of $15$ are $1,3,5$ and $6=5+1$ is not a power of $2$.