Let $H_1,H_2$ be two subgroups of $G$. Show that $ G=H_1 \cup H_2 \implies G=H_1 \lor G=H_2 $

Question

This is my third week of abstract algebra.

Let $H_1,H_2$ be two subgroups of $G$. Show that $$ G=H_1 \cup H_2 \implies G=H_1 \lor G=H_2 $$

Here is what I thought:

If we consider subgroups of $\Bbb Z/4\Bbb Z=\{0,1,2,3 \}$, then this rule would be true:

\begin{equation} \forall H_1,H_2\subseteq G :H_1\subseteq H_2 \lor H_2\subseteq H_1 \tag{1} \end{equation}

Which would prove the statement. However, if I look at the vectorspace $\Bbb R^3$, then $(1)$ wouldn't be true. For that example, one of those subgroups has to be three dimensional, does there exist something as dimensions in group theory ?

I think I'm looking in the wrong direction, can somebody enlighten me a little bit here ? A subtle hint would be appreciated.

Answer 1

Hint: Suppose $\,a_1\in H_1-H_2\;\;,\;\;a_2\in H_2-H_1\,$, then: where is $\,a_1a_2\,$ ? So what can we deduce?

Answer 2

Thanks for the hints, is this proof correct?

Suppose $G=H_1 \lor G=H_2$ is not true.

This implies $G\not= H_1 \land G\not=H_2$, and this implies that $H_1,H_2\subsetneq G$.
So there exists an element $h_1 \in G$ that's not in $H_2$. And because $G=H_1 \cup H_2$, this element must be in $H_1$. Same goes for an element $h_2 \in G$ which is in $H_2$ but not in $H_1$. We now have that there exist $h_1\in H_1 - H_2\text{ and }h_2\in H_2-H_1$.

We now consider $h_1\cdot h_2$. As $G=H_1\cup H_2$, this element must be either in $H_1$ or $H_2$. Take the case $h_1\cdot h_2\in H_1$ (the other case is similar). We know that since $h_1 \in H_1$, $h_1^{-1}\in H_1$. But then also $h_1^{-1}\cdot h_1\cdot h_2 \in H_1$. This implies $h_2\in H_1$, but this contradicts what we already showed: $h_2\in H_2-H_1$. $\square$

Answer 3

First, you can't prove the theorem by consider subgroups in one group such as $\mathbb Z_4$ (for that matter you won't be proving the theorem even if you manually checked for a trillion cases). You need a general argument.

To structure your proof, use proof by contradiction. Assume that the desired result does not hold and try to find a contradiction. Thus, start by saying "assume that $H_1,H_2\subset G$ are subgroups of $G$ such that $H_1\cup H_2$ is a subgroup but neither $H_1\subseteq H_2$ nor $H_2\subseteq H_1$."

Now, these assumptions give you something. Namely, elements $h_1\in H_1$ and $h_2\in H_2$ such that $h_1\notin H_2$ and $h_2\notin H_1$. Now remember what it means to be a subgroup and ponder about what may happen.