Proving S4 is generated by a 2-cycle and a 3-cycle

  • Thread starter Thread starter lus1450
  • Start date Start date
lus1450
Messages
40
Reaction score
1

Homework Statement


Show ##S_4## (symmetric group on ##4## letters) can be generated by two elements ##x## and ##y## such that ##x^2 = y^3 = (xy)^4##


Homework Equations





The Attempt at a Solution


I'm guessing I can use ##(12)## and ##(143)##. I got this since I know ##S_n = \langle (12), (13), (14) \rangle## by theorem in my text, and ##(13)(14) = (143)##. I know by another theorem that ##S_n = \langle (12),(1234) \rangle \text{ and } (1234) = (143)(143)(12)##, so I believe that means I am allowed to say they both generate the same group. Furthermore, ##(12)(143) = (1432)##, which is a ##4##-cycle that will be the identity when raised to the fourth. I know here it's not written rigorously, but that is my mindset. Is this correct? Thanks in advance
 
Physics news on Phys.org
It looks good to me.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top