Finding 2-Sylow & 3-Sylow of ##S_4##

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Homework Statement



Find the three 2-Sylow subgroups of ##S_3## and find a 2-Sylow subgroup and a 3-Sylow subgroup of ##S_4.##

2. The attempt at a solution

I got ##|S_3| = 6 = 2\dot\ 3## and ##|S_4| = 24 = 2^3\dot\ 3.## So ##S_3## has a a Sylow 2 subgroup of order 2 and a Sylow 3 subgroup of order 3. I am asked to find the three 2-Sylow subgroups of ##S_3## so since the 2-Sylow of ##S_3## has order 2 is it just the permutations ##(1 2), (1 3), (2 3)? ##
 
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Lee33 said:

Homework Statement



Find the three 2-Sylow subgroups of ##S_3## and find a 2-Sylow subgroup and a 3-Sylow subgroup of ##S_4.##

2. The attempt at a solution

I got ##|S_3| = 6 = 2\dot\ 3## and ##|S_4| = 24 = 2^3\dot\ 3.## So ##S_3## has a a Sylow 2 subgroup of order 2 and a Sylow 3 subgroup of order 3. I am asked to find the three 2-Sylow subgroups of ##S_3## so since the 2-Sylow of ##S_3## has order 2 is it just the permutations ##(1 2), (1 3), (2 3)? ##

Those are the generators of the only subgroups of S_3 of order 2, so yes.
 
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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...
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