As a specific example, let us classify groups of order 21. Let $G$ be a group of order 21. There is only one Sylow 7-subgroup $K$ so it is normal. The possibility of there being conjugate Sylow 3-subgroups is not ruled out. Let $x$ denote a generator for $K$ , and $y$ a generator for one of the Sylow 3-subgroups $H$ . Then $x^7=y^3=1$ , and $yxy^{-1}=x^i$ for some $i<7$ since $K$ is normal. Now $x=y^3 x y^{-3}=y^2 x^i y^{-2}= y x^{i^2} y^{-1}=x^{i^3}$ , or $i^3=1 \mod 7$ . This implies $i=1,2$ , or 4.

Case 1: $yxy^{-1}=x$ , so $G$ is abelian and isomorphic to $C_{21}=C_3 \times C_7$ .

Case 2: $yxy^{-1}=x^2$ , then every product of the elements $x,y$ can be reduced to one in the form $x^i y^j$ , $0 \leq i <7$ , $0 \leq j <3$ . These 21 elements are clearly distinct, so $G=\inn{x,y \mid x^7=y^3=1, yx=x^2y}$ .

Case 3: $yxy^{-1}=x^4$ , then since $y^2$ is also a generator of $H$ and $y^2 x y^{-2}=y x^4 y^{-1}=x^{16}=x^2$ , we have recovered case 2 above.