The subgroup intersection criterion makes sure that any set intersection of subgroups of a group $G$ is also its subgroup. But how about the set union of subgroups? The following counterexample shows that it is not true in general:

Consider the two subgroups $$\begin{array}{rcl}H_1&:=\{2n\mid n\in\mathbb Z\}=\{\ldots,-6,-4,-2,0,2,4,6,\ldots\}\\

H_2&:=\{3n\mid n\in\mathbb Z\}=\{\ldots,-9,-6,-3,0,2,6,9,\ldots\}\end{array}$$ of the additive group $(\mathbb Z,+)$ of all integers. The union $$H_1\cup H_2=\{\ldots,-9,-8,-6,-4,-3,-2,0,2,3,4,6,9,\ldots\}$$ cannot be a group (and more than ever a subgroup) because it is not closed under the addition operation. For instance, $$-3+2=1\not\in H_1\cup H_2.$$

| | | | created: 2020-06-27 12:53:48 | modified: 2020-06-27 12:54:38 | by: *bookofproofs* | references: [8311]

[8311] **Modler, F.; Kreh, M.**: “Tutorium Analysis 1 und Lineare Algebra 1”, Springer Spektrum, 2018, 4. Auflage