# Thread: Isomorphism between S3 and D3

1. ## Isomorphism between S3 and D3

Hi,

Someone told me that a good exercise to do to get better understanding of groups is proving that S3 and D3 (the dihedral group of order 6) are isomorphic. Is there any other way to do that than explicitly constructing a function from one group to the other and show that it is a bijective homomorphism or writing out the multiplication tables and seeing they match?

I remarked that each group as one element of order 1 (the identity), 2 elements of order 3 and 3 elements of order 2. Is this sufficient to show that the groups are isomorphic i.e would it be possible to find a group of order 6 with exactly the same number of elements with these orders that is not isomorphic to S3 or D3?

2. If you want to understand this at an intuitive, geometric level, you should think of both groups as the group of symmetries of an equilateral triangle.

Any symmetry of the triangle can be described by the way that it permutes the vertices, and any permutation of the vertices can be achieved by a symmetry. So the group of symmetries is isomorphic to S3.

Slightly less obviously, the group of symmetries is generated by two elements, a rotation through $2\pi/3$ radians about the centre, and a reflection in a line joining a vertex to the midpoint of the opposite side. These two operations satisfy the defining relations for the generators of D3.