Find a group and three proper subgroups such that their union equals the original group. Symbolically: Find a group $G$ with proper subgroups $H,K,L$ such that
$$G = H \cup K \cup L.$$
Since set unions don't respect group operations, it's hard for me to find subgroups that "close themselves" under the group operation when they're unionized. Any hints?
EDIT: Furthermore, how is this supposed to be feasible when we have that unions of subgroups are not groups unless the constituent sets are comparable? (Perhaps I am reading the statement incorrectly?)
 A: The Klein 4-group $V_4$ is the union of three proper subgroups. 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. 
Theorem (Bruckheimer, Bryan and Muir) A group is the union of three proper subgroups if and only if it has a quotient isomorphic to $V_4$.
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. $5, (2009)$.   
