Isomorphism between $D_3$ and $S_3$

Question

I ran into the contradiction when I was trying to explicitly show that $D_6$ and $S_3$ are isomorphic following the outline given in my textbook. I couldn't resolve it, despite spending considerable amount of time. I also couldn't find the answer to my question here.

I am showing the contradiction I have stumbled upon in the following image:

EDIT:

I made a simple mistake in the image above. The first permutation should be $(321)$. Then everything checks out and there is no contradiction.

Answer 1

Actually, you just made a simple mistake. Your first permutation (the rotation) should be $$ \begin{bmatrix}1&2&3\\3&1&2\end{bmatrix} = (132) $$ With this, the equation checks out.

So you shouldn't think what you wrote in a comment: "I incorrectly assumed that every symmetry of equilateral triangle should correspond to a geometrically motivated permutation." Actually you were right from the beginning. The isomorphisms between $D_6$ and $S_3$ are given naturally by the way the geometric transformations permute the numbered corners. ("Isomorphisms" in plural because we have freedom in the way we number the corners.)

Of course, to formally prove they are isomorphic, you need to get a bit more technical, like in the other answers. A way to formalize this conceptual argument is to note/show that $D_6$ acts faithfully on the three corners. This implies that the action gives rise to an injective homomorphism $D_6\to S_3$, and since the groups are of equal finite order, it's an isomorphism.

Edit: Woops, now I’m the one who messed up! I fixed the cycle notation, should be fine now.

Answer 2

Note that the dihedral group $D_3$ is given by $$\{1,a,a^2,b,ab,a^2b\}$$ where $a$ is understood to be the rotation over $120$ degrees and $b$ is the 'flip' over a diagonal.

Similarly, $S_3$ is given by $$\{Id,(123),(132),(12),(13),(23)\}.$$

Let $\phi\colon D_3\to S_3$ be the map determined by $\phi(a)=(123)$ and $\phi(b)=(12)$. Note that if we want $\phi$ to be a group homomorphism, we must set $\phi(a^ib^j)=\phi(a)^i\phi(b)^j$. I leave it to you to verify that $\phi$ yields a well-defined group homomorphism. Furthermore, it is then easy to show that $\phi$ is either injective or surjective (once you show one, the other automatically follows as $\phi$ is map from a set with 6 elements to a set with 6 elements).

Also, note that I used the shorter cycle notation for permutation. I recommend looking up this shorthand notation if you have not seen it before.

Answer 3

There's an indirect way to show that $S_3$ and $D_3$ are isomorphic which is not generalizable and somewhat lengthy but, imho, still worth pointing to at a student beginning a journey in group theory.

By working directly with the multiplication tables is straightforward to construct by hand groups of small order. For instance, it is rather trivial to show that there is only one group of order $2$, $3$ or $5$ which turn out to be cyclic. Also one easily reckons the possibility of two groups of order $4$, both commutative.

The analysis of the possibile situations involving 6 elements is slightly more complicated but eventually one should convince oneself that there are only two possibilities: a cyclic group with $6$ elements and a non commutative group $G_6$, the smallest possible non-commutative group.

Now $D_3$ and $S_3$ have $6$ elements each and are non-commutative, so they must both be instances of this group $G_6$ and so they must be isomorphic.