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I'm trying to write a different proof from the book I'm using since I am seeing a connection to this corollary from the same book. However I am not certain if the connection I see is correct.

Corollary: Let $a$ have order $k$ modulo n, then $a^h$ has order k if and only if $gcd(h,k)=1$.

Proof of the title of this post:

Suppose $d|(p-1)$ and $a^d\equiv 1\pmod p$ such that order of $a$ is $d$ modulo p. Thus $\forall i, 1\leq i\leq d-1$, $a^i \not\equiv 1\pmod p$. By the corollary, There exists $\phi(d)$ $a^i$'s such that the order of $a^i$ modulo p = $d$. Since every $a^i \equiv k\pmod p$ such that $1\leq k\leq p-1$. Then by a bijective relation between all the powers of $a$ and $k$, there exist $\phi(d)$ integers between and including $1$ and $p-1$ having order $d$ modulo p.

  • See https://math.stackexchange.com/questions/2303826/if-g-is-a-primitive-root-modulo-m-and-k-m-1 – lab bhattacharjee Nov 22 '17 at 16:07
  • is my proof incorrect? – TheLast Cipher Nov 22 '17 at 16:08
  • You start by assuming that there is some $a$ such that $a^d\equiv 1\pmod p$ and $a^k\not\equiv 1\pmod p$ for all $1\le k< d$ (in other words, that for any $d\mid p-1$, an element of order $d$ in $(\Bbb Z/p\Bbb Z)^*$ actually exists). This is not so easily proved and it is, in fact, the tricky part of what you wish to prove. –  Nov 22 '17 at 16:18
  • So it's not guaranteed that there exist such $a$ from $1$ to $p-1$ such that $p$ is prime? – TheLast Cipher Nov 22 '17 at 16:22
  • I have not said that. I've said that, unless you already know that, fixed your prime $p$, there is some $a$ such that all the integers coprime with $p$ are congruent to some power of $a$, you are not by any means allowed to make the assumption in the beginning of your proof. –  Nov 22 '17 at 16:25
  • So I first need to prove that every power of $a$ is congruent to a unique integer $k$, $1\leq k\leq p-1$? – TheLast Cipher Nov 22 '17 at 16:30
  • But isn't that assumption a part of the condition of the conditional statement I'm trying to prove? – TheLast Cipher Nov 22 '17 at 16:33

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