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.


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?

  • $\begingroup$ aaaaah thanks ! $\endgroup$ – Kasper Feb 25 '13 at 19:38

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: $h_2\in H_2-H_1$. $\square$


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.