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.

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    $\begingroup$ 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”. $\endgroup$ May 4, 2021 at 2:29
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    $\begingroup$ The result is true if you replace “proper” with “proper nontrivial”, so the distinction in this situation is immaterial. $\endgroup$ May 4, 2021 at 2:30
  • $\begingroup$ 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. $\endgroup$
    – Mike
    May 4, 2021 at 2:37
  • $\begingroup$ @Mike: That’s not a standard reading or a standard interpretation of the statement. $\endgroup$ May 4, 2021 at 3:22
  • $\begingroup$ 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. $\endgroup$ May 4, 2021 at 16:51

1 Answer 1


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

  • $\begingroup$ This is a very interesting approach and something I, ironically, thought of immediately after. But is the proof I wrote not correct? $\endgroup$
    – JeremyS
    May 4, 2021 at 2:12
  • $\begingroup$ Your proof is correct. $\endgroup$
    – user403337
    May 4, 2021 at 2:17
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    $\begingroup$ @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. $\endgroup$
    – Erick Wong
    May 4, 2021 at 2:35
  • $\begingroup$ @ErickWong it's a useful general fact $\endgroup$
    – user403337
    May 4, 2021 at 2:41

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