I want to show $A_n$ can be defined by the following relations $x_1,x_2,\dots,x_{n-2}$: $ x_1^3;\qquad x_i^2, i>1;\qquad (x_ix_{i+1})^3;\qquad (x_ix_j)^2, j>i+1. $ I assume $n\geq 3$, and let $x_1=(123)$, $x_i=(12)(i+1,i+2)$ for $2\leq i\leq n-2$. These satisfy the given relations, so $A_n$ is the homomorphic image of $FG^{(n-2)}/K$, where $K$ is the normal subgroup generated by the realtions and $FG^{(n-2)}$ the free group generated by $x_1,\dots,x_{n-2}$.
Now $x_1^3\in K$, and so $x_1^3=1$ in $FG^{(1)}/K$, thus $|FG^{(1)}/K|\leq \frac{3!}{2}=3$. By induction I assume $ |FG^{(n-3)}/K|\leq \frac{(n-1)!}{2}. $ If $H$ is the subgroup of $FG^{(n-2)}/K$ generated by $x_1,\dots,x_{n-3}$, I claim $ FG^{(n-2)}/K=H\cup Hx_{n-2}\cup Hx_{n-2}x_{n-3}\cup\cdots\cup Hx_{n-2}\cdots x_2x_1\cup Hx_{n-2}\cdots x_2x_1^2. $ This follows since the last relation implies $x_ix_jx_ix_j=1$, or $x_ix_j=x_jx_i$ for $j>i+1$, and the cubed relation implies $ x_{n-3}x_{n-2}x_{n-3}=x_{n-2}x_{n-3}x_{n-2}. $ So for any $g\in FG^{(n-2)}/K$, I can write $g=hx_{n-2}x_{j_1}\cdots x_{j_m}$, for $h\in H$. So if $j_1\neq {n-3}$, I can commute to the left of $x_{n-1}$. If $j_1=n-3$, then I look to $j_2$, if $j_2=n-2$, then I can use the second relation above to decrease the number of elements to the right of the first $x_{n-2}$, etc. Either way, it is clear that $g$ has is in one of the cosets of $H$ listed above. So $ |FG^{(n-2)}/K|\leq (n-1)\frac{(n-1)!}{2}+\frac{(n-1)!}{2}=\frac{n!}{2}. $
This would show $A_n\simeq FG^{(n-2)}/K$ if the homomorphism is surjective, since I would have an epimorphism from a group of order $\leq n!/2$ onto a group of order $n!/2$. Is the induced homomorphism surjective? I know $A_n$ is generated by $3$-cycles , and even just cycles of form $(1ij)$, but I can't figure a way to get these with the $x_i$ I chose.
Source: Jacobson's Basic Algebra I, pp. 71, #5.