Peculiar presentation of Symmetric group of degree 10

Question

In the context of some problem I am working on, I got this peculiar presentation of a group. I have established computationally that this group is $S_{10}$, but I was wondering if it can be done algebraically? $$ \langle s_1,s_2,s_3,s_4,s_5,s_6|s_i^2, s_1s_6s_1^{-1}s_6^{-1}, s_2s_5s_2^{-1}s_5^{-1}, s_3s_4s_3^{-1}s_4^{-1}, (s_1s_2)^6,(s_1s_3)^6,(s_1s_4)^6,(s_1s_5)^6,(s_2s_3)^6,(s_2s_4)^6,(s_2s_6)^6, (s_3s_5)^6,(s_3s_6)^6,(s_4s_5)^6,(s_4s_6)^6,(s_5s_6)^6\rangle $$ In words, I have six order-2 generators, each of which commutes with exactly one other generator. If they do not commute then the order of $s_is_j$ is 6.
Are there any tricks that will allow me to pull this off?

Answer 1

The presentation you've given isn't $S_{10}$. It isn't finite; you can just make words like $s_1 s_2 s_3$ of infinite order.

As you said any time $[s_i,s_j]\not= 1$ you have that $o(s_is_j)=6$, but if you want to use that you still need to define $[s_i,s_j]$ for the other pairs of elements.

Note that there are generating sets of $S_{10}$ for which those relations hold. For example: $$\begin{align*}s_1&=(16)(27)(38)(49)(5\text{X})\\ s_2&=(12)(56)\\ s_3&=(12)(36)\\ s_4&=(16)(23)(47)(5\text{X})\\ s_5&=(12)(37)\\ s_6&=(12)(67) \end{align*}$$ but of course these obey additional relations which would be necessary to define a group presentation of $S_{10}$ associated with these generators.