Using the universal property of the free group, I want to show that $$S_3 \cong G = \langle a,b: a^3=b^2=e;ba=a^2b \rangle.$$
I think I understand the general idea of how to show that a group is isomorphic to its presentation in the form of a relation on a free group. However, when I try to execute such proof, something seems off, though I'm not sure what. In particular, showing that $\phi$ (defined below) is an injection (which, as I understand, should be done by using the relation on G to find a limit to the order of G) seems off to me though I can't seem to understand why. I'd appreciate any corrections or comments on the details of my argument or of the validity of the argument in general.
First, write $S_3 = \{ (1), \beta = (12), (13), (23), \alpha = (123), (132) \}$. If we define a map of sets $f: \{ a,b\} \rightarrow S_3$ such that $f(a) = \alpha$ and $f(b) = \beta$, then, by the universal property of the free group, there exists a group homomorphism $\phi: F({a,b}) \rightarrow S_3 $ such that $\phi \circ \iota = f$. The map $\phi(x_1 … x_n) = f(a_1)^{\epsilon_1} … f(a_n)^{\epsilon_n}$, where $x_i = a_i^{\epsilon_i}$, $a_i \in \{ a, b \}$ and $\epsilon_i \in \{ -1, 1 \}$. We have that $\phi$ is a surjective homomorphism since $\{ \alpha, \beta \}$ generates $S_3$.
To show that $\phi$ is an injection, we need only prove that $|G| \leq 6$. Given our relation, we can write that, for any word $w \in \langle a, b \rangle$, $\hat{w} = \hat{b}^i \hat{a}^j$, w/ $i \in \{ 1 , 2 \}$, $j \in \{ 0 , 1 , 2 \}$ since $\hat{b}^2, \hat{a}^3$. So $|G| = |\{1,2 \} \times \{ 1,2,3 \}| = 6$.
Therefore, $\phi:G \rightarrow S_3$ is an isomorphism, so we are done.
Best Answer
You $\phi: F(a,b)\to S_3$ can't be an injection since $|F(a,b)|=\infty$.
Since $\phi$ respects the relations of $G$, by von Dyck, it factors through $G=\frac{F(a,b)}{\langle\langle a^3=b^2=bab^{-1}a^{-2}=1\rangle\rangle}$ hence you have
$\bar{\phi}: G=\frac{F(a,b)}{\langle\langle a^3=b^2=bab^{-1}a^{-2}=1\rangle\rangle}\to S_3$ which is still a surjection. Then try to show that this is an injection