The presentation $$\langle s_3,\dots, s_n|s_i^3=1, (s_is_j)^2=1, 3\le i\neq j\le n\rangle$$ is valid for alternating groups of order $n$. I have to prove this. (This is Exercise 7.2 of Oleg Bogopolski's book Introduction to group theory, on its page 64.)
One approach suggested in the text is
$ s_i\to (1 2 i)$.
I am unable to establish that it is a presentation by this method.
Can there be a way to establish equivalence with a standard presentation?
Best Answer
I expect you are intended to mimic the proof of Theorem 7.1, which states that the standard Coxeter presentation of $S_n$ on $n-1$ generators really does present $S_n$.
Lets call the group defined by the presentation $G_n$. The assignment $s_i \mapsto (1,2,i)$ extends to a homomorphism $G \to A_n$, which is surjective, so $|G_n| \ge |A_n|$, and we just need to prove that $|A_n| \le |G_n|$.
We do that by induction on $n$, and the base case $n=3$ is easy.
For $n>3$, define $H$ to be the subgroup $\langle s_3, s_4,\ldots, s_{n-1} \rangle$ of $G_n$. Then, by induction, $|H| \le |A_{n-1}|$, so we need to prove that $|G_n:H| \le n$.
For that, we need to guess $n$ (right) coset representatives of $H$ in $G_n$, and then prove that the guess is correct. Note that the image of $H$ under the homomorphism is the stabilizer of $n$ in $A_n$. The trick in making the guess is to choose words whose images are coset representatives of this stabilizer. That is, words for which the image of $n$ is $i$ for $1 \le i \le n$.
For example, we could choose words $$g_1,g_2,\ldots,g_n = s_n, s_n^{-1},s_n^{-1}s_3,\ldots,s_n^{-1}s_{n-1}.$$ (I compose permutations from left to right.)
Now we have to prove that, for any coset rep $g_i$ and any generator $s_j$, we have $g_is_j \in \cup_{i=1}^n Hg_i.$
I am not going to do that in detail, but here are a few examples with $n=6$.
$s_6s_4 = s_4^{-1}s_6^{-1}$; $s_6^{-1}s_3s_4 = s_6^{-1}s_4^{-1}s_3^{-1} = s_4s_6s_3^{-1} = s_4s_6s_3ss_3 = s_4s_3^{-1}s_6^{-1}s_3$; $s_6^{-1}s_3s_6 = s_6s_3^{-1} = s_3^{-1}s_6^{-1}s_3$.