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Suppose, I am given two elements and asked to show that they generate a given subgroup of a group, how can I show that ? I have read about the closures but don't know how to apply them in problems. For example, I found two questions, one asks to show that $(1234)$ and $(1243) $ generates $S_4$, and the other one was involving matrices where I have to show that two $2\times 2$ matrices one having the $ a_12 $ position $0 $ and the others $1$, while the other one has the $ a_21 $ position $0$ and the others $1$, they generate special linear group of $2\times 2 $ matrices over $\mathbb{Z}_3$ which is a subgroup of the whole group $G$ of $2\times 2$ matrices.

Thank you in advance .

raven chaser
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    Not sure there is a general theory of such things...keep in mind that the unsolvability of the Word Problem makes it unlikely that there will be a simple, "one-size-fits-all" approach. Best to take each situation as it arises. – lulu Sep 14 '18 at 20:41
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    Is $(1234)$ and $(1432)$ a typo or unusual notation? If it is cyclic notation then $(1234)=(1432)^{-1}$ so the subgroup they generate is $\langle(1234)\rangle\ne S_4$ – Robert Chamberlain Sep 15 '18 at 07:39
  • The closest to a generic method that immediately springs to mind is coset enumeration. After checking the elements are indeed in the group (hard in general, but easy in the given examples), one needs only check that the subgroup generated by these elements has index $1$. – Robert Chamberlain Sep 15 '18 at 07:46

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For the symmetric group $S_n$ there are several sets of generators known, e.g. $(12\cdots n)$ and $(12)$, or many others. The proofs give some insight how to deal with this questions. Keith Conrad has a note on Generating Sets, where several such results are given (see below).

Reference: Generators of $S_n$

Conrad's notes here; Example $1.8$, page $2$.

Dietrich Burde
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  • I will go through that note and try to understand it. They are using the same basis like concept similar to linear algebra, for making such groups. This note looks pretty interesting and thank you for the reference. I am grateful. – raven chaser Sep 15 '18 at 09:25