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.

Sorry, this is not an already asked question as Moishe marked it. I reopened it — DonAntonio, Feb 02 '18 at 22:06

score 3 · Accepted Answer · answered Feb 02 '18 at 21:41

3

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.

answered Feb 02 '18 at 21:41

Akababa

3,109

Why $;H_3\not\subset H_1;$ implies the existence of those two elements $;h_1,,h_2;$ in $;H_3;$ ? – DonAntonio Feb 02 '18 at 21:56
It implies it has an element not in $H_1$, but since it's in $H_1\cup H_2$ it has to be in $h_2\in H_2$ Similar thing for $h_1$. – Akababa Feb 02 '18 at 21:57
I agree with the first part: there exists $;h_2\in H_3\setminus H_1;$ , and since $;h_2\in H_1\cup H_2;$ , then in fact $;h_2\in H_2\setminus H_1;$ , but why the other one too? Why there is $;h_1\in H_3;$ s.t. $;h_1\in H_1\setminus H_2;$ ? – DonAntonio Feb 02 '18 at 22:05
It's in the question: $H_3 \subseteq H_1 \cup H_2$ and $H_3 \not\subset H_2$ – Akababa Feb 02 '18 at 22:06
Sorry, then I meant otherwise in my past comment: it is clear there exists $;h_2\in H_3\setminus H_1\subset H_1\cup H_2;$ , so then $;h_1\in H_2\setminus H_1;$ , but why the other one too? – DonAntonio Feb 02 '18 at 22:09
$H_3 \not\subset H_2\implies \exists h_1\in H_3\backslash H_2\subseteq (H_1\cup H_2)\backslash H_2=H_1\backslash H_2$ – Akababa Feb 02 '18 at 22:11

score 0 · Answer 2 · answered Feb 02 '18 at 21:50

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 $H_3 \subseteq H_2$.

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 $H_3 \subseteq H_1$.

2 Answers2