# What (finite?) groups are covered by 3 (disjoint?) proper subgroups? [duplicate]

$$C_2^2$$ is disjointly* covered by $$3$$ proper subgroups. I couldn't find any other group that is disjointly covered by $$3$$ proper subgroups. The groups $$C_2\times C_{2n}$$ for $$n\geq 0$$ (i define $$C_0 = \mathbb Z$$) are all covered by $$3$$ proper subgroups, but not disjointly so unless $$n=1$$. This can be verified by looking at the subgroups generated by $$(1,0), (0,1)$$ and $$(1,1)$$, and it will be a disjoint cover iff $$n=1$$.

Are there other (finite?) groups which are (disjointly?) covered by $$3$$ proper subgroups? Has there been a classification of them?

*by disjoint i mean up to the identity element

• There is a classification: the group must have $C_2\times C_2$ as a quotient, and this is sufficient. I'll dig up a reference later today. Commented Aug 2 at 19:48
• What do you mean by "disjointly?" Subgroups must all include $1,$ so they can't be disjoint. Commented Aug 2 at 21:09
• @ThomasAndrews See the final line in my question.. Commented Aug 2 at 21:33

This is answered in Bhargava's paper Groups as Unions of Proper Subgroups. There it is shown that a group is the union of $$3$$ proper subgroups if and only if it has $$C_2\times C_2$$ as a quotient. No finiteness assumption is necessary.
The proof shows that this union is "disjoint" precisely when the kernel is trivial, so the group is isomorphic to $$C_2\times C_2$$.