Why in a finite field with subfields such $|G| $ divides $|H|$ but $G \not\subset H$?

Question

In a finite field $F$ of order $n$, $F^*$ under multiplication is isomorphic to $Z_{n-1}$. So, if we have subfields $G$ and $H$ such that $|G| $ divides $|H|$, then by the fundamental theorem of cyclic groups, we should have $G \subset H$, since $G$ and $H$ are subgroups of $F^*$ under multiplication. But, I know this not true. Can someone tell what is wrong with my argument?

Answer 1

Consider $\mathbb F_{p^n}$. You can show that the subfields of $\mathbb F_{p^n}$ are of the form $\mathbb F_{p^m}$ for $m | n$. Moreover, there is exactly one copy of $\mathbb F_{p^m}$ inside $\mathbb F_{p^n}$ for each $m$.

This fits nicely with your description of the multiplicative groups, since a cyclic group of order $p^m - 1$ is certainly a subgroup of a cyclic group of $p^n - 1$ if $m | n$. To spell it out, if $\sigma$ is a generator of the multiplicative group of $\mathbb F_{p^n}$, then $\sigma^{(p^n - 1)/(p^m - 1)}$ is a generator of the multiplicative group of $\mathbb F_{p^m}$.

Now take two subfields $\mathbb F_{p^{m_1}}$ and $\mathbb F_{p^{m_2}} $ in $\mathbb F_{p^n}$. Their intersection is a copy of $\mathbb F_{p^{{\rm gcd}(m_1,m_2)}}$.

Think about the multiplicative groups of the three subfields. The multiplicative group of $\mathbb F_{p^{m_1}}$ is generated by $\sigma^{(p^n - 1)/(p^{m_1} - 1)}$. The multiplicative group of $\mathbb F_{p^{m_2}}$ is generated by $\sigma^{(p^n - 1)/(p^{m_2} - 1)}$. The intersection of these two multiplicative groups is generated by $\sigma^{(p^n - 1)/{\rm gcd}(p^{m_1} - 1,p^{m_2} - 1)}$, which is equal to $\sigma^{(p^n - 1)/(p^{{\rm gcd}(m_1,m_2)} - 1)}$. And this is precisely the generator for the multiplicative group of $\mathbb F_{p^{{\rm gcd}(m_1,m_2)}}$.

Answer 2

I think I could have guessed what saravanan meant: let $\;F_{p^k}\;$ be a finite field, so we know that

$$F_{p^k}^*:=F_{p^k}\setminus\{0\}\;\;\text{ is a multiplicative cyclic group of order}\;\;p^k-1$$

and thus in fact $\;F_{p^k}^*\cong C_{p^k-1}\cong\Bbb Z_{p^k-1}=\;$ the cyclic group of order $\;p^k-1\;$ .

Now, if $\;G,H\;$ are subfields of $\;F_{p^k}\;$, then we know that $\;G=F_{p^m},\,H=F_{p^n}\;$ , with $\;m,n\,\mid\,k\;$ , and thus we also have that

$$G^*,\,H^*\le F_{p^k}^k\implies p^m-1,\,p^n-1\,\mid\,(p^k-1)$$

which is also true because we know that $\;m\,\mid\,k\iff p^m-1\,\mid p^k-1\;$ , and the proof of this isn't long.

Answer 3

Fundamental theorem of cyclic group cannot be applied to $G$ and $H$, since they need not form a group under multiplication. It can be only applied to $G^*$ and $H^*$