using $|g^k|= \frac{|g|}{gcd(|g|,k)}$
give a necessary and sufficient condition for when $[a] ∈ Z/nZ $ has order n. How many elements [x] ∈ Z/nZ have the property that ⟨[x]⟩ = Z/nZ?
so I know the order of a = smallest positive k such that $a^k=e$ and we want a condition in that $a^n=e$ and I have also proved previously that $|g^k|= \frac{|g|}{gcd(|g|,k)}$ is true
I'm having trouble wrapping my head around this, any help would be great
No elements in $\mathbb{Z}/n\mathbb{Z}$ have order $n$, where the operation is multiplication, as $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is a multiplicative group with order strictly less than $n$.
And if you are asking instead how many elements generate $(\mathbb{Z}/n\mathbb{Z})^{\times}$, not every group of the form $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is cyclic i.e., has a single generator either. Take $n=8$.
IF the operation is addition, then every element $a$ s.t. $(a,n)=1$ has order $n$. Indeed suppose not. Then $ma = kn$ for some positive $m < a$. This implies that $n|ma$. If $n$ does not divide $m$ then this implies that gcd$(a,n) > 1$. [Indeed, that $n$ does not divide $m$ implies that for some prime $p$ and some positive integer $e$, that $p^e|n$ but $p^e$ does not divide $m$. But that $n|ma$ implies that $p^e$ divides $ma$ which implies that $p$ must divide $a$ which implies gcd$(n,a) > 1$.
