Can a group be a union of three subgroups? I am being asked to show an example of when this fails or prove it rigorously.  I am thinking of using an example to disprove the claim that a group can be a union of three subgroups.  However I am not so sure if proving it rigorously might be a better way.  What do you guys think?  
 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. $5, (2009)$.   
A: this may not be much help, but there is a numeric constraint which might serve as a filter (if my understanding of the question  is correct)
if the group is of finite order $g$, the subgroups $g_1,g_2,g_3$, their intersections $g_{12},g_{23},g_{31},g_{123}$ then we require:
$$
g-g_1-g_2-g_3+g_{12}+g_{23}+g_{31} - g_{123} = 0
$$
subject to the constraints of the division lattice implied by Lagrange's theorem.
the quaternion group gives:
$$
8 -4-4-4+2+2+2-2 =0
$$
