Find the order of the element g^n by already knowing the order of g 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 


*

*$\text{ord}(g^n) \mid \frac{d}{\text{gcd}(n,d)}$

*$d \mid n \cdot \text{ord}(g^n) \implies \frac{d}{\text{gcd}(n,d)} \mid \text{ord}(g^n)$
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.
 A: 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 $e = \gcd(n,d)$. 
A: 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 $d = \text{ord}(g) \mid n \cdot \text{ord}(g^n)$.
A: 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 $ord(g^n) =\frac d{\gcd(n,d)} = \frac {ord (g)}{\gcd(n,d)}$.
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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 $a|dm\implies a|m$.
