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The solution uses the following proof which I don’t fully understand:

Let $d:=\text{ord}(g)$

Because $$(g^n)^{d/\text{gcd}(n,d)} = (g^d)^{n/\text{gcd}(n,d)}=e$$ it follows that

  1. $\text{ord}(g^n) \mid \frac{d}{\text{gcd}(n,d)}$
  2. $d \mid n \cdot \text{ord}(g^n) \implies \frac{d}{\text{gcd}(n,d)} \mid \text{ord}(g^n)$

and it becomes clear that $\text{ord}(g^n)= \frac{d}{\text{gcd}(n,d)}$.

I have a problem understanding the second point and would really appreciate a hint or two.

Sverre
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Pg1234
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  • Welcome to Math SE. I assume your "bcd" is a typo for "gcd" and, as mentioned in my answer, I assume that in the first part of $2$ that $ord(g)$ should be $ord(g^n)$ instead. – John Omielan Feb 18 '20 at 21:26
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    $d|n\cdot ord(g^n)\implies \frac d{\gcd(n,d)}|\frac n{\gcd(n,d)}\cdot ord(g^n)$. But $\frac d{\gcd(n,d)}$ and $\frac n{\gcd(n,d)}$ are relatively prime. So $\frac d{\gcd(n,d)}|\frac n{\gcd(n,d)}\cdot ord(g^n)\implies \frac d{\gcd(n,d)}|ord(g^n)$. – fleablood Feb 18 '20 at 21:39

3 Answers3

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Consider in general

$$a \mid bc \tag{1}\label{eq1A}$$

with

$$e = \gcd(a,b) \implies a = ef, \; b = eg, \; \gcd(f,g) = 1 \tag{2}\label{eq2A}$$

Then \eqref{eq1A} becomes

$$ef \mid egc \implies f \mid gc \implies f \mid c \tag{3}\label{eq3A}$$

with the last implication due to $\gcd(f,g) = 1$. Thus, applying this to the $2$ part in your question, you get what they show. In particular, you have $a = d$, $b = n$, $c = \text{ord}(g^n)$ (I assume missing that power of $n$ is a typo), and $e = \gcd(n,d)$.

John Omielan
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2

Since $\frac{d}{\text{gcd}(n, d)} \mid d$, we have $\frac{d}{\text{gcd}(n, d)} \mid n \cdot \text{ord}(g^n)$, and since $\frac{d}{\text{gcd}(n, d)}$ and $n$ are coprime it follows that $\frac{d}{\text{gcd}(n, d)} \mid \text{ord}(g^n)$.

Also, $d \mid n \cdot \text{ord}(g^n)$ indeed holds, since $g^{n \cdot \text{ord}(g^n)} = (g^n)^{\text{ord}(g^n)} = e$, and so $d = \text{ord}(g) \mid n \cdot \text{ord}(g^n)$.

Sverre
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2

The second point is just:

$a|n*b\implies \frac ak|\frac {nb}k$ and $\frac a{\gcd(a,n)}|\frac n{\gcd(a,n)}b$. But as $\frac a{\gcd(a,n)}$ and $\frac n{\gcd(a,n)}$ are relatively prime we will always have $a|n*b\implies \frac a{\gcd(a,n)}|b$.

So if we ever have both

1) $ord(g^n)|\frac d{\gcd(n,d)}$ and

2) $d|n*ord(g^n)$ we will have

2a) $\frac d{\gcd(n,d)}|ord(g^n)$.

And as $a|b$ and $b|a$ can only occur (assuming positive values) if $a = b$

And if we have both 1) and 2a) we have $ord(g^n) =\frac d{\gcd(n,d)}$.

So that's it.

....

We have, by a simple theorem, that if $a^m = e$ then $ord(a)|m$

And, by definition we have, $e = (g^n)^{ord(g^n)} = g^{n*ord(g^n)}$.

So that means we have 2): $d=ord(g) |n*ord(g^n)$.

And as we have 2) we have 2a).

And we can get 1) from the following:

We have $(g^n)^{\frac d{\gcd(n,d)}} = g^{n\cdot \frac d{\gcd(n,d)}}= g^{\frac n{\gcd(n,d)}\cdot d} = (g^d)^{\frac n{\gcd(n,d)}}$.

And as $d = ord(g)$, $g^d = e$ and $ (g^d)^{\frac n{\gcd(n,d)}}=e^{\frac n{\gcd(n,d)}}=e$.

So we have 1). $ord(g^n)|\frac d{\gcd(n,d)}$

And that's that.

We have 2a) and 1) so $ord(g^n) =\frac d{\gcd(n,d)} = \frac {ord (g)}{\gcd(n,d)}$.

=======

In this post I assume two very basic Lemmas:

A) $\frac a{\gcd(a,b)}$ and $\frac b{\gcd(a,b)}$ are always relatively prime.

B) If $a^k = e$ then $ord (a)|k$.

.... oh.... and guess a third

C) If $a$ and $d$ are relatively prime tha $a|dm\implies a|m$.

fleablood
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