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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?

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    $\begingroup$ You "know that is not true"? For example...? $\endgroup$ – DonAntonio Feb 27 '17 at 11:06
  • $\begingroup$ I don't have any example but a book says that if $|G|=p^n$ and $|H|=p^m$, their intersection is a field of order $p^s$ with $s=gcd(n,m)$ $\endgroup$ – Saravanan Feb 27 '17 at 11:11
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    $\begingroup$ But if $\;F_{p^n}\;,\;\;F_{p^m}\;$ are fields, then $\;F_{p^m}\le F_{p^n}\iff m\,\mid\,n\;$ Are you taking this into consideration? And what has to do what that book says with what you say in your question? I don't really see it clearly and I think you should write down a good explanation on this. $\endgroup$ – DonAntonio Feb 27 '17 at 11:15
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    $\begingroup$ If $|G^*|$ divides $|H^*|$ then, indeed, $G\subseteq H$ by cyclicity of $F^*$. However, it may easily happen that $|G|\mid |H|$ but $|G^*|\nmid |H^*|$. The smallest example is when $F=\Bbb{F}_{64}$. It has a subfield $G$ of order $8$ and another subfield $H$ of order $4$. Here $(8-1)\mid (64-1)$ and $(4-1)\mid (64-1)$ as we should. But $(4-1)\nmid (8-1)$ meaning that $H$ cannot be a subfield of $G$. We easily see that $G\cap H=\{0,1\}$ is the prime subfield with only two elements. $\endgroup$ – Jyrki Lahtonen Feb 27 '17 at 12:12
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    $\begingroup$ In other words, your error is that you forgot that only the multiplicative group of non-zero elements of a finite field is cyclic. You need to leave out zero, and therefore you need $|G|-1$ to divide $|H|-1$. The additive group of $F$ is not cyclic (unless $n$ is a prime, when it has no proper subfields), so you cannot similarly argue with the orders of additive groups. $\endgroup$ – Jyrki Lahtonen Feb 27 '17 at 12:21
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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 $\mathbb F_{p^{{\rm gcd}(m_1,m_2)}}$.

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  • $\begingroup$ but looking $F_{p^m}$ as a multiplicative group makes it a subset of $F_{p^n}$, right? $\endgroup$ – Saravanan Feb 27 '17 at 11:32
  • $\begingroup$ Are you trying to show that $\mathbb F_{p^m}$ is a subfield of $\mathbb F_{p^n}$? Yes, $\mathbb F_{p^m}$ is a subset of $\mathbb F_{p^n}$, and yes, they have the same multiplicative properties. But you also need them to have the same addivitive properties! $\endgroup$ – Kenny Wong Feb 27 '17 at 11:36
  • $\begingroup$ "if $|G|=p^n$ and $|H|=p^m$ their intersection is a field of order $p^s$ with $s=gcd(n,m)$", isn't the statement contradicting the statement $F_{p^m}$ subset $F_{p^n}$ $\endgroup$ – Saravanan Feb 27 '17 at 11:42
  • $\begingroup$ And if $m | n$, then $s = gcd(n,m) = m$. So the intersection is the whole of $F_{p^m}$. $\endgroup$ – Kenny Wong Feb 27 '17 at 11:44
  • $\begingroup$ my question is different, see the question again, i am taking $F_{p^n}$ and $F_{p^m}$ are subfields of some bigger finite field, let say orders is $p^{100}$ with n=50 and m=20 $\endgroup$ – Saravanan Feb 27 '17 at 11:47
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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.

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  • $\begingroup$ yeah absolutely, thanks, Jyrki Lahtonen comment on my answer clarifies the problem $\endgroup$ – Saravanan Feb 27 '17 at 12:32
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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 $H^*$

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