For two subgroups $H_1, H_2$, how can I show $H_1 \cup H_2$ is subgroup $\iff$ $H_1 \subseteq H_2$ or $H_2 \subseteq H_1$

Question

I want to shot that for two subgroups $H_1, H_2$, $H_1 \cup H_2$ is a subgroup iff $H_1 \subseteq H_2$ or $H_2 \subseteq H_1$

The implication $\impliedby$ follows easily, as (without loss of generality) if $H_1 \subseteq H_2$ then $H_1 \cup H_2 = H_2$ which is a subgroup by assumption.

It's easy to give counterexamples for the case where (wolg) $H_1 \not \subseteq H_2$, e.g. $2\mathbb{Z}$ and $3\mathbb{Z}$, but I'm struggling with giving a rigorous proof for either $(\lnot \impliedby \lnot)$ or $(\implies)$

I'm really stuck, and would appreciate any hints.

Answer 1

Hint: If neither is subgroup contained in the other, then they each contain an element that's not contained in the other. Consider the product of those two elements.