Let $G$ be an abelian group. Suppose $H = \left \{g \in G\ \mid\ \text {either}\ \text {ord} (g)\ \text {is infinite or}\ g = e \right \}$ is a subgroup of $G.$ Then show that either all the elements of $G \setminus \{e \}$ are of infinite order or all the elements of $G$ are finite.
Let us suppose that not all the elements of $G$ are of finite order. Let $T$ denote the torsion subgroup of $G.$ Then by the given hypothesis we have $T \subsetneq G.$ We need to prove that order of all the elements of $G \setminus \{e\}$ are infinite. Suppose in contrary $\exists$ $a \in G \setminus \{e\}$ such that $\text {ord} (a) < + \infty.$ Then $T$ contains a non-trivial element $a.$ Now since $G \setminus T \neq \varnothing$ let $x \in G \setminus T.$ Then $ax \notin T$ for otherwise since $a \in T$ we have $x \in T,$ a contradiction. So $ax \in H.$ But this implies $a \in H,$ since $x \in H.$ But that means $a=e$ (since $\text {ord} (a) < +\infty$), a contradiction.
QED
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