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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?

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    To get a presentation of the abelianization of a group, just take an existing presentation and throw in relations saying that every pair of generators commutes. Using the commutation relations first, we get that the abelianization of the given group is $\mathbb{Z}^6/(2 e_i, (6 e_i + 6 e_j)_{i+j \neq 7})$, where $e_i$ are the generators of $\mathbb{Z}^6$ and I have switched to additive notation to emphasize that the group is abelian. The first set of relations implies the second, so we have $(\mathbb{Z}/2)^6$.2012-12-16

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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.