Why do we have $S_3 \times C_2 = D_6$ ?
Visualize $D_3$ as a hexagon, and $S_3$ as its three diagonals. ![[Hexagon.svg]]S1 ![[lines.svg]]S2 There is something subtle here. We see each straight line as a whole and don't care about the points on the lines. So when we say its "symmetry group" we are talking about the permutation group of a 3-element set. More rigorously, we first list all the permutation in regard to the points on the shape and then identify some of them as we need. When doing this we must show and only need to show that the permutations are all still well-defined, i.e. the elements in the set we identified must be "inseparable". As elements in the same equivalence class, they must be mapped to the same equivalence class. And the hexagon is equivalent with ![[lines 1.svg]]S3 Where we have arrows rather than lines, i.e. we have six elements in the set that we want to permute. So S1 < S2 (we denote the corresponding group of the shape by S1, S2, S3). Apparently S1 = S3 > S2 Comparing S2 and S3, the additional element can be described as central symmetry (rotation by 180 degree). This element, surprisingly, lies in $Z(D_6)$ . That explains why it's a direct product factor.