# Proof that $S_3$ isomorphic to $D_3$

So I'm asked to prove that $$S_{3}\cong D_{3}$$ where $D_3$ is the dihedral group of order $6$. I know how to exhibit the isomorphism and verify every one of the $6^{2}$ pairs, but that seems so long and tedious, I'm not sure my fingers can withstand the brute force. Is there a better way?

• The dihedral group is always a subgroup of the corresponding symmetric group since it permutes the vertices of a polygon. Since they have the same size in this case, they must be equal. – Cheerful Parsnip Sep 15 '14 at 21:44

Every symmetry of the triangle permutes its vertices, so there is an embedding $D_3\to S_3$ already just from geometry. Then it suffices to check they have the same order.

• realy nice answer. – mesel Sep 15 '14 at 21:44

Method 1

The most direct way is to use a presentation for the groups. You know a general dihedral group of order $2n$ has a presentation:

$$D_{n}=\langle a,b | a^2=b^n=e, aba=b^{-1}\rangle$$

Then just establish that with $n=3$, and by choosing $a=(12)$ (or any transposition) and $b=(123)$ (or $(321)$) that the corresponding elements of the symmetric group satisfy the same relations, and hence are the same group under the isomorphism

$$\begin{cases} D_{3}\to S_3 \\ a\mapsto (12) \\ b\mapsto (123)\end{cases}$$

Method 2

If you're not familiar with group presentations, you can note it geometrically, since any permutation of the three vertices of a triangle gives rise to a distinct symmetry of that shape, so that there is an injective homomorphism from $S_3\to D_3$, and since they have the same order, it is an isomorphism.

Method 3

You can consider that $S_3/C_3\cong D_3/C_3$ where here the $C_3$ are the unique subgroups of order $3$, and then show that the only possibilities for groups with such a quotient group are either direct or semi-direct products, and since neither is abelian, they must be isomorphic.

Method 4

You can write $D_3=\{e,a,b,b^2, ab, ab^2\}$, and then verify by looking at the Cayley table that

$$\begin{cases}a\mapsto (12) \\ b\mapsto (123)\end{cases}$$

is an isomorphism. This is the most tedious way, but when all else fails.

Hint: $S_3=\langle(1,2,3),(1,2)\rangle$ and $D_3=\langle a,b\rangle$ set $\phi(a)=(1,2,3)$ and $\phi(b)=(1,2)$ i.e send generaters to generaters.

Note: Choose $a$ from the ones of order $3$ and b from the ones of order $2$.