Why mismatch in finding number of elements of given order in cyclic group.

Question

It is stated here :

In a cyclic group of order $n$, the order of $a^k = \frac{n}{\gcd(n,k)}.$ Further, the distinct elements that have order $n/d$ are : $a^{di}: i\in\mathbb{Z_{n/d}^{\times}}$

So, for $n=8$, check for different values of order $n/d= 8, 4, 2,1$:

$|a^1|= 8/1= 8,$
$|a^2|= 8/2= 4,$
$|a^4|= 8/4= 2,$
$|a^8|= 8/8= 1,$

The distinct elements that have order $8/1= 8$ are :
$a^{i}: i\in\mathbb{Z_{8}^{\times}}$

The distinct elements that have order $8/2= 4$ are :
$a^{2i}: i\in\mathbb{Z_{4}^{\times}}$

The distinct elements that have order $8/4= 2$ are :
$a^{4i}: i\in\mathbb{Z_{2}^{\times}}$

The distinct elements that have order $8/8= 1$ are :
$a^{i}: i\in\mathbb{Z_{1}^{\times}}$

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The above shows one element of order $8$, but there are two: $a,a^7$.

The set generated under multiplication operation by $\langle a \rangle =\{a,a^2,a^3,a^4,a^5,a^6, a^7,e\},$
The set generated under multiplication operation by $\langle a^7 \rangle = \{a^7,a^6,a^5,a^4,a^3,a^2, a^1, e\}.$

Answer 1

Note that $a, a^3, a^5, a^7$ are generators of $G=\{e,a,a^2,...,a^7\}$, because they are coprimes of 8.

Answer 2

Your formula says the elements of order $8$ are $a^i$ for $i\in\mathbb Z_8^\times$. Since $\mathbb Z_8^\times=\{1,3,5,7\}$, this means the elements of order $8$ are $a, a^3, a^5$ and $a^7$, which is correct.