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Prove $n\mid d \iff a^d-1\mid a^n-1$ and prove field with $p^d$ elements is a subfield of a field with $p^n$ elements $\iff$ $d\mid n$

So for the first part i'm betting there is some elementary algebra identities that would help a lot, and then the second part will follow from the first part because the splitting field of $p^n-1$ over $\mathbb{Q}$ is a field with $p^n$ elements but if anyone wants to talk about what's going on here in more depth that'd be cool. Thanks!

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    $\begingroup$ I suppose your intial hypothesis is $d\mid n$, not $n\mid d$. $\endgroup$ – Bernard Nov 1 '18 at 23:21
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Here $$a^n-1=(a^d-1)(a^{n-d}+ a^{n-2d}+ \cdots +a^d+1)=(a^d-1)k$$ so $a^d-1$ divides $a^n-1$ whenever $d$ divides $n$.


$\Longleftarrow$:

Let $F$ be a field with $p^n$ elements and assume $d$ divides $n$. Now consider $$F_1=\{x \in F: x^{p^d}=x\}$$ Check $F_1$ is a sub field of $F$! We have $\vert F_1 \vert$ is atmost $p^d$, since $x^{p^d}=x$ has atmost $p^d$ roots in $F$. Since multiplicative group of $F$ is cyclic, we have $\langle F^*\rangle=\langle a\rangle$. Using this to see $a^k \in F_1$ an so $F_1$ is a subfield of order $p^d$. Moreover $F_1$ is unique(?)

$\Longrightarrow$: Suppose $F_1$ is a sub field of order $p^d$ in a field $F$ of order $p^n$. Now $$n=[F: \Bbb{F}_p]=[F:F_1][F_1:\Bbb{F}_p]=[F:F_1] \cdot d$$ so $d$ divides $n$


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  • $\begingroup$ What is $k$ and why is $a^k \in F_1$? $\endgroup$ – Math is hard Nov 2 '18 at 18:57
  • $\begingroup$ I already mention $k$ in the first equation, namely, $k=a^{n-d}+\cdots+1$. That is we know $a^d-1$ divides $a^n-1$, so $a^n-1=(a^d-1).k, k \in \Bbb{Z}$ . I mention this $k$ $\endgroup$ – Chinnapparaj R Nov 3 '18 at 2:39
  • $\begingroup$ how does the fact that you wrote $a^d-1$ as a divisor of $a^n-1$ depend on the fact that $d$ divides $n$? $\endgroup$ – Math is hard Nov 3 '18 at 16:46
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Hint:

You have the high-school identity: $$a^k-1=(a-1)(a^{k-1}+a^{k-2}+\dots+a+1).$$

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