Prove/disprove: Let $G$ be abelian and $n\in\Bbb N$. Then $\{ g\in G:|g| \leq n\} $ forms a subgroup of $G$ [closed]

Let $G$ be an abelian group, and $n \in \mathbb{N}$. Prove or disprove: $\left\{ g\in G:\left|g\right| \leq n\right\} $ forms a subgroup of $G$.

I'm pretty sure it's false, and the group of all roots of unity would be a counterexample, but I'm not sure.

Consider $2,3 \in \mathbb{Z}_6$ (of orders $3$ and $2$ respectively). Then, $2+3=5$ has order $6$, which gives a counterexample for $G=\mathbb{Z}_6$ and $n=3$. Of course, this could be converted into an example inside the group of all roots of unity by considering $e^{2{\pi}i/3}=(e^{{2\pi}i/6})^2$ and $-1=(e^{{2\pi}i/6})^3$.