I need to find the order of non-isomorphic subfields of GF(64). According to Lagrange's theorem, the order of a subfield has to divide the order of the "superfield". $64 = 2^6$ so the order of our subfield has to be in $\{1,2,4,8,16,32,64\}$. The Fields with order 1 and 64 are two non-isomorphic subfields. According to the question there needs to be two more but I can't decide..

For two fields to be isomorphic to each other, there needs to be a bijection between them, so they have to have the same cardinality. As far as I can see, none of $\{2,4,8,16,32\}$ equal to 64 so how do I need to proceed here?

Might it be that a Field with order 32 is not necessarily a subfield of GF(64) for example?

thank you for your help


The general answer is this:

Let $p$ be a prime number. The field $\mathbf F_{p^m}$ is (isomorphic to) a subfield of $\mathbf F_{p^n}$ if and only if $m\mid n$.

  • $\begingroup$ I didn't find that in my textbook but thanks. the only thing I can find is that the order of the subfield has to divide the order of the superfield. A proof of your Theorem/Lemma would be much appreciated :) $\endgroup$ – DariusTheGreat Jul 7 '17 at 9:49
  • $\begingroup$ It is generally proved with ‘existence and uniqueness. A proof is based on the description of a finite field as the splitting field of some polynomial $X^{p^n}-X$ and the uniqueness of a splitting field. See for instance Wikipedia. $\endgroup$ – Bernard Jul 7 '17 at 10:02

You have applied Lagrange's theorem for the additive group of $GF(64)$.

Now apply it to the multiplicative group, which has $63$ elements, to conclude:

  • $GF(64)$ cannot contain a subfield of order $32$ or $16$ because neither $31$ nor $15$ divide $63$.

  • $GF(64)$ can contain a subfield of order $8$, $4$, $2$ because $7$, $3$, $1$ divide $63$.

You still need to prove existence of subfields of order $8$, $4$, $2$. Lagrange's theorem shows a way.

For instance, a subfield of order $8$ would be the set of elements in $GF(64)$ such that $x^8-x=x(x^7-1)=0$. Since $GF(64)$ is a field, there are at most $8$ elements in this set. However, it is not easy to see that this set has any elements other than $0$ and $1$.

The simplest solution is to prove that the multiplicative group of a finite field is cyclic. Then it follows that $x^7-1=0$ has exactly $7$ solutions in $GF(64)$ and so the subfield of order $8$ is the set of solutions of $x^8-x$. This set is clearly closed under multiplication. That it is closed under addition follows from the Frobenius identity, $(a+b)^2 = a^2 + b^2$.

  • $\begingroup$ but aren't the subfields with order 8,4,2 isomorphic to GF(64) according to Bernard(the answer below)? I concluded from his answer that the subfields have order 16,32 and 1. $\endgroup$ – DariusTheGreat Jul 7 '17 at 11:42
  • $\begingroup$ @user3047143, what are the divisors of $6$? $\endgroup$ – lhf Jul 7 '17 at 11:43
  • $\begingroup$ 1,2,3 and according to below subfields of order 2,4,8 are isomorphic which are the one's that we're not looking for. $\endgroup$ – DariusTheGreat Jul 7 '17 at 11:45
  • $\begingroup$ @user3047143, I think you need to reread Bernard's answer. $\endgroup$ – lhf Jul 7 '17 at 11:50
  • $\begingroup$ oh my bad i calculated $n|m$....stupid mistake by me $\endgroup$ – DariusTheGreat Jul 7 '17 at 11:52

In my opinion, the simplest proof of the theorem cited by @Bernard is to apply naturally Galois theory to the Galois extension $\mathbf F_ {p^n}/\mathbf F _p$. Its Galois group $G$ is cyclic of order $n$, generated by the Frobenius automorphism $Fr$ defined by $Fr (x)=x^p$. The subextensions correspond by Galois to the subgoups of $G$, which are exactly the groups of order $m$ for all $m$ dividing $n$, generated by powers of $Fr$ .


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