How many subfields does a field with $2^{36} = 68719476736$ elements have

Question

I encounter to this qustion. This is such a big number so I believe there must be some trick to solve it.

But I wonder what property of finite fields should I use. I know that according to Galois correspondence. The subfields are correspond to divisor of $2^{36}$. Is that correct? So is are there exactly $37$ such subgroups?

Thanks for any help!

Answer 1

You may have established that $G \cong \mathbb{Z}/36\mathbb{Z}$ (since you have a finite extension of finite fields).

By the converse of Lagrange's theorem (for cyclic groups), there exists subgroups $H$ which have order dividing the order of the group, namely, $|H| =2,3,4,9,6$. This is all of them (don't forget the trivial ones).

Using the fundamental theorem of Galois theory you should use that bijection you were talking about.