Let $G$ be a cyclic group and $T(G)=\{a\in G:\text{ord}(a)<\infty \}$ the group of the torsion-elements. I need to find $T(G)$ for $G$ finite or infinite. Here is what I tried, ist this correct?
i) $(a)=G$ finite, then $\exists n \in \mathbb{N}, n<\infty:\text{ord}(a)=|G|=n$ and $a^n=e$. Since for every $x\in G$ it holds that $x^{|G|}=e$, then also every element of $G$ is in $T(G)$.
ii) $(a)=G$ infinite, then $G\cong\mathbb{Z}$. For every $x\in G, x\neq e, (x)$ is a (cyclic) subgroup. Every subgroup of $\mathbb{Z}$ has the form $m\mathbb{Z}$ for a positive (if $x=e, m=0$) $m$ and $|m\mathbb{Z}|=\infty$, so only $e\in G$ is in $T(G)$.
//Edit: with the hints in comments (Thank you!)
Since $G$ is cyclic, it is $1$-generated. Let $a$ be a generator. If $a$ has finite order, then $a \in T(G)$ and so $T(G)=G$. If $a$ has infinite order, then so does any other nontrivial element of $G$ (since it can be written as $a^k$ for some $k \in \mathbb{Z}$). In particular then $T(G)=\{1\}$.