 Solution

The intersection $H\cap K$ contains the identity element. If $a,b\in H\cap K$, then $a{b}^{-1}$ belongs to both $H$ and $K$, so it belongs to their intersection. The subgroup test proves that $H\cap K$ is a subgroup of $G$, with the inherited operation. Explicitly, closure and inverses come from this test, and associativity is inherited from $G$.

Label the vertices of the equilateral triangle $1,2,3$ counterclockwise. For the dihedral group of order six, let $r=(123)$ be the counterclockwise rotation by $120^{\circ }$, and let $s=(23)$ be reflection in the line through vertex 1 and the opposite side's midpoint. ChooseThen $B$ rotates by $240^{\circ }$. The three Euclidean reflections $C,D,E$ fix vertices $1,3,2$ respectively. Products act on a vertex from right to left. The relations $r^3=s^2=e$ and $sr=r^{-1}s$ givewhere exponents of $r$ are reduced modulo three and those of $s$ modulo two. This constructs the entire Cayley table:For example $AC=D$ whereas $CA=E$, so the group is nonabelian.

A subgroup of order two is generated by its unique nonidentity element, which must have order two. The rotations $A,B$ have order three, and each reflection has order two. Thus there are exactlyNone is a normal subgroup. Conjugation by $A$ sends the reflection $r^{j}s$ to $r^{j+2}s$, cycling the three reflections. For example $AC{A}^{-1}=E\notin ⟨C⟩$. Each of the other two subgroups fails the same conjugation test.

Finally, take two distinct such subgroups. For instance their union is $⟨C⟩\cup ⟨D⟩=\{I,C,D\}$, but $CD=B$ lies outside it. Thus it is not closed under the inherited group operation and is not a group. The distinctness is necessary: the union of a subgroup with itself is that subgroup.