# No group can be a union of two proper subgroups [duplicate]

I am trying to prove the following result.

No group $$G$$ can be the union of two proper subgroups.

The first point of confusion is that I have two different definitions of proper subgroup. My professor defined it as a subgroup $$H \subsetneqq G$$, while the textbook (Artin) defines it as a subgroup $$H \subset G$$ that is neither $$\{e\}$$ or $$G$$. The former seems more standard to me. Is that correct? I'm going to stick with that in the below attempt.

Suppose for the sake of contradiction that $$G = H \cup K$$ for subgroups $$H,K \subsetneqq G$$. Then $$H \not \subset K$$ and $$K \not \subset H$$ because, otherwise, $$H \cup K = K$$ and $$H \cup K = H$$, respectively, which contradicts $$K$$ and $$H$$ being proper subgroups. So there exists $$x \in G \setminus H$$ and $$y \in G \setminus K$$. So $$xy \in G$$, so $$xy \in H$$ or $$xy \in K$$. If $$xy \in H$$, since $$y$$ and hence $$y^{-1}$$ are. elements of $$H$$, we have $$(xy)y^{-1} = x \in H$$, a contradiction. Similarly, if $$xy \in K$$, since $$x$$ and hence $$x^{-1}$$ are elements of $$k$$, we have $$x^{-1} (xy) = y \in K$$, a contradiction. So this construction is impossible.

• Artin’s definition is not usual; one speaks of a proper subgroup of $G$ as a subgroup that does not equal $G$; and of a nontrivial subgroup of $G$ as one that does not equal $\{e\}$. Artin’s definition is what one normally calls “a proper nontrivial subgroup”. May 4, 2021 at 2:29
• The result is true if you replace “proper” with “proper nontrivial”, so the distinction in this situation is immaterial. May 4, 2021 at 2:30
• Your proof looks good. Meanwhile here "proper" subgroups $H$ and $K$ must mean in this context that neither $H$ nor $K$ is a subset of the other.
– Mike
May 4, 2021 at 2:37
• @Mike: That’s not a standard reading or a standard interpretation of the statement. May 4, 2021 at 3:22
• Compare also your proof with the standard proofs from the duplicates, e.g., this one, or this one. It shows how to make the proof shorter and more elegant. As a byproduct you can verify if your proof is correct or not. May 4, 2021 at 16:51

## 1 Answer

Hint: Prove that for the union of two subgroups to be a group, they must be nested.

• This is a very interesting approach and something I, ironically, thought of immediately after. But is the proof I wrote not correct? May 4, 2021 at 2:12
• Your proof is correct.
– user403337
May 4, 2021 at 2:17
• @JeremyS Your proof is essentially a rearrangement of this approach (you show that the union of non-nested groups is not a group). So I’m not sure why this answer written as a hint. May 4, 2021 at 2:35
• @ErickWong it's a useful general fact
– user403337
May 4, 2021 at 2:41