If a union of two subgroups of $G$ equals $G$, must one of the subgroups equal $G$?

If the union of two subgroups of $$G$$ is the group $$G$$, does that mean one subgroup is $$G$$?

In a problem, it was proven to be true on this site. But take real numbers under addition as a group with Rational numbers as a subgroup and irrational numbers with $${0}$$ as a subgroup. Then their union is a real number, but no subgroup is real numbers. So the statement is not true for all groups, right?

• Irrational numbers under addition as a subgroup of $\Bbb R$? Are you sure of this? – Juniven Mar 8 '17 at 2:51
• under addition with 0 and negative irrational nubers. – jnyan Mar 8 '17 at 2:52
• Recall that the identity must be in the subgroup. What can you say to $0$? Is it rational or irrational? – Juniven Mar 8 '17 at 2:54

The result that one of the subgroups must be $G$ is correct. What goes wrong with your example is that the irrational numbers together with $0$ are not a subgroup of $\mathbb{R}$. For instance, $\pi$ and $1-\pi$ are both in this set, but their sum $\pi+(1-\pi)=1$ is not.

• so any union of proper subgroups cannot be $G$? – jnyan Mar 8 '17 at 2:58
• That is correct (assuming you are taking a union of only two proper subgroups). – Eric Wofsey Mar 8 '17 at 3:05
• what if more than 2? – jnyan Mar 8 '17 at 3:21
• got it. klien 4 group – jnyan Mar 8 '17 at 3:22

The irrational numbers with $0$ do not form a subgroup of $\mathbb{R}$. $(\sqrt{2}+1)-\sqrt{2}=1$, so the irrationals with $0$ aren't closed under subtraction.

You're correct that at least one of them must be $G$. This follows from the more general fact that if the union of two groups is a group, then one of the groups must be (isomorphic to) a subgroup of the other.

The decimal-part of an irrational number has a complement which together add to 1. Thus the integers are in any group of irrational numbers. SO your example doesn't work and you are false.

• lol and +1 at "you are false" – tilper Mar 8 '17 at 2:55