If $H_{3}$ is subgroup of $G$ that satisfies $H_3 \subseteq H_1 \cup H_2$ and $H_3 \not\subset H_2$, then $H_3 \subseteq H_1$.

Question

Let $(G,*)$ be group and $H_{1}, H_{2}$ subgroups of $G$. If $H_{3}$ is subgroup of $G$ that satisfies $H_3 \subseteq H_1 \cup H_2$ and $H_3 \not\subset H_2$, then $H_3 \subseteq H_1$. Prove or give counterexample.

I really don't know how to solve this ,any help would be appreciated.

Answer 1

Let's prove it by contradiction:

Assume $H_3\not\subseteq H_1$, so $H_3$ has two elements $h_1\in H_1\backslash H_2, h_2\in H_2\backslash H_1$. Then $h_1+h_2\in H_3\subseteq H_1\cup H_2\implies h_1+h_2\in H_1$ or $h_1+h_2\in H_2$.

If $h_1+h_2\in H_1$, then $h_2=(h_1+h_2)-h_1\in H_1$, contradiction. Similarly $h_1+h_2\in H_2$ leads to a contradiction.

Answer 2

It is well known and easy to prove that a group cannot be the union of two proper subgroups. And $H_3 \subseteq H_1 \cup H_2 \iff H_3=(H_3 \cap H_1) \cup (H_3 \cap H_2)$. Hence $H_3=H_3 \cap H_1$ or $H_3=H_3 \cap H_2$, that is $H_3 \subseteq H_1$ or $H_3 \subseteq H_2$.