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Please help or hints me to solve this question:

Prove: if $F$ is a finite field, then $H \cup \{0\}$ is a subfield of $F$ for every subgroup $H$ of the multiplicative group $F^*$ if and only if the order of $F^*$ is either $1$ or a prime number of the form $2^p - 1$ with a prime $p$.

Masoud
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  • $\mathbf{F}{p^n}$ is a field with $p^n$ elements and $\mathbf{F}{p^n}^$ is a cyclic group with $p^n-1$ elements. What are the subgroups of $\mathbf{F}_{p^n}^$, the subfields of $\mathbf{F}_{p^n}$, what if $n$ is prime, and if $p^n-1$ is prime ? – reuns Jul 23 '17 at 06:21
  • @ reuns, The order of subgroup $F^_{p^n}$ is divided by $p^n-1$, The order of subfield $H$ of $F^_{p^n}$ is $p^s$ where $s|n$. – Masoud Jul 23 '17 at 06:39
  • You mean $N | p^n-1$ and $p^s$ and $s | n$. What if $n$ is prime, what if $p^n-1$ is prime ? – reuns Jul 23 '17 at 06:41
  • @Arthur: Yes, Thanks. – Masoud Jul 23 '17 at 08:26
  • @ reuns, I can not prove that $p=2$ in $p^n -1$. Please help me. – Masoud Jul 23 '17 at 15:24
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    $p$ is prime. What happens if it is odd ? What are the subgroups of $\mathbf{F}_5^*$ ? – reuns Jul 23 '17 at 17:48
  • Because $F^*$ is cyclic, what must happen is that all the divisors of $p^n-1$ must be of the form $p^m-1$ where $m\mid n$. That is a tall order in most cases. – Jyrki Lahtonen Jul 23 '17 at 19:09
  • The answer can be found in the following question. https://math.stackexchange.com/questions/2980751/finite-field-where-each-subgroup-of-the-multiplicative-group-of-nonzero-elements – RuhAllah37 Jan 18 '24 at 07:20

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