# Difficulty in proving that S_3 is isomorphic to the free group on two letters with the following relation:

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.

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