Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
The presentation seems garbled and the obvious fix makes it the trivial group. –  i. m. soloveichik Dec 16 '12 at 13:47
2  
$s_3 s_6 s_1^{-1} s_4^{-1}$ is this a typo? –  Billy Dec 16 '12 at 13:49
2  
To me the obvious fix is to change $s_3 s_6 s_1^{-1} s_4^{-1}$ to $s_3 s_4 s_3^{-1} s_4^{-1}$. I'm not sure what group that is, but it is neither trivial nor $S_{10}$, as the abelianization is $(\mathbb{Z}/2)^6$. –  David Speyer Dec 16 '12 at 14:26
    
@DavidSpeyer, how can you be that sure what the group's abelianization is without actually find out what the group itself is? Is there some method/trick you're using here? –  DonAntonio Dec 16 '12 at 14:50
2  
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 $(\mathbb{Z}/2)^6$. –  David Speyer Dec 16 '12 at 15:22
add comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.