I'm trying to understand a proof that $D_3 \cong S_3$. I could write out the multiplication tables, which would work, but I think arguing geometrically is the most instructive and what my instructor intended. The sketch I have in my mind is as follows:
Every symmetry of the equilateral triangle permutes the vertices, so $D_3$ can be regarded as a subgroup of $S_3$, hence there exists an injective homomorphism $f: D_3 \to S_3$. As $|D_3| = 6 = |S_3|$, any injective map of sets is also surjective, so $f$ is a bijective and therefore an isomorphism.
The argument, as I've written it, seems too hand-wavy to me. In particular, I feel that I'm making a leap by equating symmetries of the triangle with permutations, even though they in fact permute the vertices, and I can think of a permutation of a three-element set $\{1,2,3\}$ as vertices $1$, $2$, and $3$, provided that I label them on the triangle. Every symmetry surely permutes the vertices, and in the case of a triangle, every permutation yields a symmetry because every vertex is adjacent to the other two vertices, so no matter how I permute them, distances between vertices is preserved. The "homomorphism" part is what I'm really stuck with. I have an intuitive sense that the diagram of maps must commute, and I could prove by brute force, by checking all of the pairs of elements of $D_3$, that it does, but I don't have a very good way to prove that the composition laws agree.
I'd appreciate any help in parsing this.
Best Answer
If you agree that: $$D_n=\langle s,r\mid s^2=r^n=1, sr=r^{-1}s\rangle$$ and have group action at hands, then both the "homomorphism part" and the injectivity come easy. In fact, every group $G$ acts (homomorphism!) by left multiplication on the left quotient set $G/H$, for every subgroup $H\le G$, and the kernel of this action is the normal core of $H$ in $G$, namely the intersection of all the conjugates of $H$ in $G$. Now, $D_n$ has the subgroup $H:=\{1,s\}$, whose index and normal core in $G$ are $n$ and (for $n>2$) the trivial subgroup $\{1\}$, respectively. In fact, $r^{-1}Hr=\{1,r^{-1}sr\}=\{1,sr^2\}$, and $sr^2\ne s$ as soon as $n>2$ (because then $r^2\ne 1$), so that $H\cap r^{-1}Hr=\{1\}$. This suffices to ensure that the normal core is trivial and hence the action is faithful (equivalently, the homomorphism is injective).