I'm trying to understand how to use Von Dyck's theorem to prove that $S_3 \cong D_3$. I believe I have a correct sketch, but I'm very fuzzy on the details, mainly because I haven't seen free groups yet and therefore don't have a full understanding of the statement of the theorem.
My rough understanding is that if I have two group presentations for groups $G$ and $H$ and the defining relations of $G$ hold in $H$, then there exists a surjective homomorphism from $G$ to $H$. So let's take $G = D_3$ and $H = S_3$. I define
\begin{align*}
S_3 & = \left \langle x, y \mid x^2 = e = y^3 \right \rangle = \left\{\mathrm{id}, (12),(13), (23), (123), (132) \right\} \\
D_3 & = \left \langle r,s \mid r^3 = 1, \; s^2 = 1, \; sr = r^{-1} s \right \rangle = \left\{e, r, r^2, s, rs, r^2 s \right\}.
\end{align*}
To do this check, do I need to "identify" the generators of one group with another? For example, it only makes sense that I identify the generators of order $2$, $x = (12)$ and $s$, and the generators of order $3$, $(123)$ and $r$. After that, I do the direct calculation. Let $\tau = (12)$ and $\sigma = (123)$. We compute:
\begin{align*}
\sigma^3 = \tau^2 = \mathrm{id} \text{ and } \tau \sigma = (23) = \sigma^{-1} \tau.
\end{align*}
Therefore, the defining relations of $D_3$ are true in $S_3$, so by von Dyck's theorem, there exists a surjective hommorphism $D_3 \twoheadrightarrow S_3$. As $|D_3| = 6 = |S_3|$, this surjective map must also be injective, and we have $D_3 \cong S_3$.
Is this the correct way to apply the theorem? Do I need to "align the generators," as I've done, or does the labelling not matter? Another way to view it is that, despite what I call the generators, I know that $S_3$ is generated by $\sigma$ and $\tau$ and I know that I can define a group by the relations I just checked, so despite what I call the elements, I've identified $S_3$ with another group. The map would ultimately map generators to generators and the "corresponding" elements to each other.
Best Answer
von Dyck's Theorem states that if you have a presentation for a group $G$ $$G = \langle X\mid R\rangle$$ (say $X=\{x_i\}_{i\in I}$, and $R=\{r_j(x_i)\}_{j\in J}$) then, whenever $K$ is any group and $k_i$ is an $I$-tuple of elements of $K$ such that for each $j$, $r_j(k_i)=e_K$, then there exists a (unique) homomorphism $\varphi\colon G\to K$ such that $\varphi(x_i)=k_i$ for each $i\in I$.
Your presentation for $S_3$ is incorrect. The presentation $$\langle x,y\mid x^2=e=y^3\rangle$$ defines the free product $C_2*C_3$, which is an infinite group. Instead, you need a relation that connects $x$ and $y$ before you get $S_3$. One example of such a presentation is: For example, take your group $$S_3 = \langle x,y\mid x^2=y^3=e, xyx=y^{-1}\rangle.$$
So, let's take the preentation I give above, $$S_3=\langle x,y\mid x^2=e=y^3, xyx=y^{-1}\rangle.$$ von Dyck's Theorem says that if you have any group $K$, and elements $g,h\in K$ such that $g^2=e$, $h^3=e$, and $ghg=h^{-1}$ all hold in $K$, then there is a homomorphism $\varphi\colon S_3\to K$ such that $\varphi(x)=g$ and $\varphi(y)=h$. Now, if you happen to know that $g$ and $h$ generate $K$, then you will also be able to conclude that $\varphi$ is surjective.
For instance, if we take the group $C_4=\{1,z,z^2,z^3\}$ the cyclic group of order $4$, I can take $g=z^2$ and $h=1$; then $g^2 = (z^2)^2 = 1$, $h^3 = (1)^3 = 1$, and $ghg = z^2(1)z^2 = z^4 = 1 = h^{-1}$. So that means that there is a homomorphism $\varphi\colon S_3\to C_3$ with $\varphi(x)=z^2$ and $\varphi(y)=1$; note that this map is not surjective.
So, we consider the presentations $$\begin{align*} S_3 &= \langle x,y\mid x^2=e=y^3, xyx=y^{-1}\rangle\\ D_3 &= \langle s,r\mid r^3=1, s^2=1, sr=r^{-1}s\rangle. \end{align*}$$ If we want to use von Dyck's Theorem to prove the groups are isomorphic, we have two options:
Option 1. To define a map from $S_3$ to $D_3$, I need to identify elements of $D_3$ that satisfy the relations satisfied by $x$ and $y$ in $S_3$. I can pick $s$ "for $x$" and $r$ "for $y$". Then $s^2=1$ and $r^3=1$ by assumption; I still need to verify that $srs=r^{-1}$ is true in $D_3$. But indeed, we know that $sr=r^{-1}s$, so $$srs = (sr)s = (r^{-1}s)s = r^{-1}(ss) = r^{-1}s^2 = r^{-1}.$$ Since $s$ and $r$ satisfy the relations satisfied by $x$ and $y$, von Dyck's theorem guarantees a homomorphism $\varphi\colon S_3\to D_3$ with $\varphi(x)=s$ and $\varphi(y)=r$.
Now, to define a map the other way, we need to find two elements in $S_3$ that satisfy the same relations that $r$ and $s$ do in $D_3$. After some thought, I pick $y$ "for $r$" and $x$ "for $s$". Then I verify that $y$ and $x$ satisfy, in $S_3$, the defining relations that $r$ and $s$ do in $D_3$. Indeed, $y^3=e$, $x^2=e$, and to verify that $xy=y^{-1}x$, we have $$xy = xy(x^2) = (xyx)x = y^{-1}x.$$ So von Dyck's Theorem guarantees the existence of a homomorphism $\psi\colon D_3\to S_3$ such that $\psi(r)=y$ and $\psi(s)=x$.
Now we have homomorphism $\varphi\colon S_3\to D_3$ and $\psi\colon D_3\to S_3$. We now note that $\psi(\varphi(x)) = \psi(s) = x$, $\psi(\varphi(y))=\psi(r) = y$. So $\psi\circ\varphi=\mathrm{id}_{S_3}$, because they agree on a generating set. And $\varphi(\psi(r)) = \varphi(y) = r$, $\varphi(\psi(s)) = \varphi(x) = s$, so $\varphi\circ\psi=\mathrm{id}_{D_3}$. Thus, $\varphi$ and $\psi$ are inverses of each other, and we have that they are isomorphisms.
Option 2. As above, we identify elements in one group that satisfy the relations that the generators satisfy in the other group. For example, $s$ "for $x$" and $r$ "for $y$". von Dyck's Theorem now gives us the homomorphism $\varphi\colon S_3\to D_3$ with $\varphi(x)=s$ and $\varphi(y)=r$.
Now, we know that $r$ and $s$ generate $D_3$, so $\varphi$ is surjective. Moreover, we know that both $S_3$ and $D_3$ have six elements each, so we have a surjective morphism from a group of order six to a group of order six. This means the map is in fact also injective, so $\varphi$ is an isomorphism, proving that $S_3$ is isomorphic to $D_3$.
Note that Option 2 requires you to know more stuff about the groups presented: you need to know their orders already.