Prove that $\mathbb{F}_{2^4}$ is not isomorphic to a subring of $\mathbb{F}_{2^5}$.
So my attempt at proving this is as follows:
Proof: Since $\mathbb{F}_{2^5}$ is finite, every subgroup has finite order, and by Lagrange's Theorem we have that any subgroup can have order $2,4,8,16$ or $32$.
Proceeding from here is my trouble. I feel as if showing that no subgroup of order 16 is exists is the only plausible way to go, but showing this is what I am unsure about. Any hints/suggestions?
Solution:
Proof: Consider the unit group of $\mathbb{F}_{2^5}$. The order of this unit group is $2^5 -1= 31$. Clearly, $31$ is prime, thus by Lagrange's Theorem we have that the only two subgroups are the trivial subgroups. Thus, adding the additive identity to the group of order one, we have $\mathbb{F}_2$, and similarly, adding the additive identity to the group of order 31, we have $\mathbb{F}_{2^5}$. Clearly, $\mathbb{F}_{2^4} \ncong \mathbb{F}_2 , \mathbb{F}_{2^5}$ since an isomorphism between finite fields can exists if they have the same number of elements.
Thank you to everyone's comments, they really helped.