I want to prove that there exists an inclusion $\mathbb{F}_{p^a} \hookrightarrow \mathbb{F}_{p^b}$ iff $a \vert b$.   Suppose that $a \vert b$, then $b =ac$ for some $c \in \mathbb{Z}$. Consider then that $p^b = p^{ac} = (p^{a})^c \implies p^a \subset p^b$. Therefore we can define an inclusion $\mathbb{F}_{p^a} \hookrightarrow \mathbb{F}_{p^b}$.   Conversely, suppose there exists an inclusion $\mathbb{F}_{p^a} \hookrightarrow \mathbb{F}_{p^b}$ and $a \ \not \vert b$. Then $b \neq ac \implies p^b \neq p^{ac} = (p^a)^c \implies p^a \not \subset p^b$ and therefore no inclusion can be defined from $\mathbb{F}_{p^a} \hookrightarrow \mathbb{F}_{p^b}$. This contradiction yields the necessary result.  Note that $p$ represents a prime.   Is this legitimate?
We know that if a finite field $F$ has characteristic $p$ (prime), then $F$ has cardinality $p^r$ where $r = [F:\mathbb{F}_p]$.   I'm now trying to say something about the possible cardinalities of subfields of $F$. I can see that there is a subfield of cardinality $p^s$ for each $s$ that divides $r$, given by the fixed field of the group generated by $\phi^s$, where $\phi$ is the Frobenius automorphism.  Now suppose $K$ is a subfield of $F$. Then (since both are additive groups), Lagrange gives us that $|K|$ divides $|F|$, so $|K| = p^t$ for some $1 \leq t \leq r $ (alternatively, $K$ contains $\mathbb{F}_p$ and so is a vector space over $\mathbb{F}_p$ and is thus isomorphic to $\mathbb{F}_p^t$, where $t = [K:\mathbb{F}_p]$). By considering the multiplicative group of units of $K$ and $F$ respectively, we get that $ p^t - 1$ divides $p^r -1$. I want to make the leap to $t|r$, but I'm failing to see why this needs to be true. Any help would be appreciated.
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