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I am doing an exercise where I am asked to prove or disprove the statement:
If $G$ is a group, and $H_1,H_2,H_3$ and $H_1 \cup H_2 \cup H_3$ are subgroups of $G$ then $\exists i,j$ with $i \neq j$ such that $H_i \subset H_j$.
In the same exercise, I've already prove that this statement is true in the case union of two subgroups is a subgroup. I have no idea how to show this statement holds for this case and I couldn't think of a counterexample. Any suggestions would be appreciated.
Hint: Klein's four element group ($\Bbb Z_2\times\Bbb Z_2$).
As you noted, a group cannot be the union of two of its proper subgroups. It might be interesting to know, that research has been done on how this might generalize. Berci gave you an example based on the Klein 4-group. In fact more is true:
Theorem (Bruckheimer, Bryan and Muir) A group is the union of three proper subgroups if and only if it has a quotient isomorphic to $C_2 \times C_2$.
The proof appeared in the American Math. Monthly $77$, no. $1 (1970)$. The theorem seems to be proved earlier by the Italian mathematician Gaetano Scorza, I gruppi che possono pensarsi come somma di tre loro sottogruppi, Boll. Un. Mat. Ital. $5 (1926), 216-218$.
For 4, 5 or 6 subgroups a similar theorem is true and the Klein 4-group is for each of the cases replaced by some finite set of groups. For 7 subgroups however, it is not true: no group can be written as a union of 7 of its proper subgroups. This was proved by Tomkinson in 1997.
There is a nice overview paper by Mira Bhargava, Groups as unions of subgroups, The American Mathematical Monthly, $116$, no. $5, (2009)$.
