Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 3 / 7D / b

If one subgroup contains the other, their union is the larger subgroup. Conversely, if neither contains the other, choose $h\in H\setminus K$ and $k\in K\setminus H$. If the union were a subgroup it would contain $hk$. If $hk\in H$, then $k=h^{-1}(hk)\in H$, a contradiction; if $hk\in K$, then $h=(hk)k^{-1}\in K$, again a contradiction. ThusThe union of two subgroups cannot generally be treated like their generated subgroup.

Choose $x\notin H$, possible because $H$ is proper. For any $h\in H$, the element $xh$ is also outside $H$, since otherwise multiplying by $h^{-1}$ would put $x$ inside $H$. Thus both $x$ and $xh$ belong to $K=⟨G\setminus H⟩$, and $h=x^{-1}(xh)\in K$. So $K$ contains $H$ as well as its entire complement, provingThe complement of a proper subgroup generates the group argument does not actually require finiteness.