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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.

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You can show that $A_n$ is generated by the permutations $x_1,\ldots,x_{n-2}$ by induction. I's true for $n=3$, and to go from $n-1$ to $n$ just show that the generated group is transitive and use the Orbit Stabilizer Theorem. –  Derek Holt May 31 '12 at 21:18
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Let $G$ be the subgroup of $A_n$ generated by $x_1,\ldots,x_{n-2}$. By inductive assumption, $x_1,\ldots,x_{n-3}$ generate $A_{n-1}$, which has order $(n-1)!/2$, and this subgroup stabilizes the point $n$. It is easy to see that, by using combinations of $x_1,\ldots,x_{n-2}$, we can map the point $n$ to any of the points $1,2,\ldots,n$. In other words $G$ acts transitively on $\{1,2,\ldots,n\}$. So the order of $G$ is $n$ times the order of the stabilizer of the point $n$; i.e. $|G| = n \times (n-1)!/2 = n!/2$, so $G=A_n$. –  Derek Holt May 31 '12 at 21:55
    
I understand now, many thanks @DerekHolt! –  Adelaide Dokras May 31 '12 at 21:59
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up vote 2 down vote accepted

The cicles $s_i = (i\ i+1)$ generate $S_n$. The function $\mathrm{sign}\colon S_n\to \mathbb{Z}_2$, with kernel $A_n$, sends every generator into $1$. So $\ker(\mathbb{sign})\cong A_n$ is generated by the products $s_is_j$.

Now $s_1s_2 = (1\ 2)(2\ 3) = (1\ 2\ 3) = x_1$. Similarly $s_1s_i = x_{i-1}$ and $s_is_j = x_{i-1}x_{j-1}$. So the set of generator of your presentation comes from the presentation of $S_n$ as a Coxeter group.

The relations of the $S_n$ presentation are $s_i^2$, $(s_is_{i+1})^3$ and $(s_is_j)^2$ in the others cases.

Let's $G$ be a group with your presentation. We can map the generator into the element $x_i = s_1s_{i+1}$ of $S_n$. From the theory of Coxeter groups we know that the function is indeed injective and well defined. So $G\cong f(G)$ where $f$ is the homomorphism defined before. But we have already proved that $f(G) = A_n$.

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