Elements $g,$ such as $g^{165}=1$ vs elements $g$ of order $165$ of a cyclic group $G$ of order $330$ I have a group $G=\langle a\rangle$ of order $330.$
I need to find all the elements $g\in G$ such as $g^{165}=1$ and then the elements $g$ of order $165.$
As far as I understand, $g\in G,g^{165}=1 \Leftrightarrow g\, \text{is} \, e,a^2,a^4,\dots,a^{330}$
And as for the elements of order $165$, those are only $e$ and $a^2$.
Am I correct or am I confusing something?
 A: Fact 1. Let $G$ be an abelian group. Then for every positive integer $n$, the set
$$
\{x\in G:x^n=1\}
$$
is a subgroup of $G$.
Fact 2. Given a cyclic group $G$ of order $m$ and a divisor $n$ of $m$, there is a unique subgroup of $G$ having order $n$.
Fact 3. Every subgroup of a cyclic group is cyclic.
Fact 4. An element of a cyclic group $G$ of order $m$ is a generator of $G$ if and only if it has order $m$.

Now you can put together facts 1–3. Since $a^2$ has order $165$, it generates a cyclic subgroup of order $165$.
This is where the elements having order $165$ live; since it is a cyclic group, you can use fact 4; set $b=a^2$ for simplicity and try to work out (or remember) what's the condition on $k$ so that $b^k$ is a generator of $\langle b\rangle$.
A: Your answer for the first question is correct. 
For the second question, $e$ has order $1$, so $e$ should not be there. And there are more. 
Question 1. Find all $g\in G$ such that $g^{165}=1$. 

Lemma 1. Let $G=\langle a\rangle$ and $|G|=n$. Let $d\mid n$. Then 
  $$\{g\in G: g^{d}=1\}=\langle a^{n/d}\rangle. $$

Proof. Let $n=qd$. 
Let $g\in LHS$. Then $g=a^{i}$ for some $i$. So $a^{id}=1$. Then $qd=n\mid id$. So $q\mid i$. Hence $a^{i}\in RHS$. Let $a^{qi}\in RHS$. Then $a^{qid}=1$. So $a^{qi}\in LHS$. 
Let $A$ be the set of all $g\in G$ such that $g^{165}=1$. Then $A=\{g\in G:g^{165}=1\}=\langle a^{2}\rangle=\{1, a^{2}, a^{4}, a^{6},\ldots, a^{328}\}$. 
Question 2. Find elements $g$ of order $165$. 

Lemma 2. Let $G=\langle a\rangle$ and $|G|=n$. Let $d\mid n$. Then 
  $$|\{g\in G: g^{d}=1\}|=d. $$

Proof. Let $n=qd$. By Dummit and Foote page 57 Proposition 5 (2), $|a^{q}|=\frac{n}{(n,q)}=d$. By Lemma 1, the result follows. 
Hence $|A|=165$. Let $B$ be the set of all $g\in G$ of order $165$. Then $B\subset A$. Since $A=\langle a^{2}\rangle$ is a cyclic group of order $165$, $B$ is the set of generators of $A$. Let $b=a^{2}$. Hence $A=\langle b\rangle$. All the generators of $A$ are elements $b^{i}$ such that $(i,165)=1$. So $B=\{a^{2i}:0\leq i\leq 165, (i,165)=1\}$. 
