Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 3 / 7D (Groups)

Use the dihedral group presentation $G=⟨r,s:r^{2n}=s^2=1,srs=r^{-1}⟩$. Every element is $r^j$ or $s{r}^j$. A nonidentity rotation has order two only when its exponent is $n$, while $(s{r}^j{)}^2=1$ for every $j$. Thus the complete list isThe half-turn $r^n$ is central, so its conjugacy class is the singleton $\{r^n\}$ and its normal closure is $\{1,r^n\}$.

For reflections, $r^k(s{r}^j)r^{-k}=s{r}^{j-2k}$ and $s(s{r}^j)s^{-1}=s{r}^{-j}$. Hence the conjugacy class is exactly the $n$ reflections whose exponents have the same parity as $j$:The subgroup generated by this class contains $r^2$, as the product of two consecutive listed reflections, and contains $s{r}^j$. Conversely those two generators contain every listed reflection. Their subgroup is invariant under conjugation by both $r$ and $s$, so the smallest normal subgroup containing $s{r}^j$ isof order $2n$ and index two. This involutions in an even dihedral group classification distinguishes the two reflection classes from the central half-turn.

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.