Generating Subgroups in <Z\stackrel{X}{13}> Modulo 13 Under Multiplication

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Homework Statement


In the group <Z\stackrel{X}{13}> of nonzero classes modulo 13 under multiplication, find the subgroup generated by \overline{3} and \overline{10}

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The Attempt at a Solution


Doesnt 3 generate {3,6,9,12} and 10 generate {2,5,10}?
 
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The problem says says "under multiplication" so the subgroup generated by \overline{3} includes all products of \overline{3}. You are adding: 3+ 3= 6, etc. 3*3= 9 and 3*9= 27= 1 (mod 13)
 
3*3 = 9 (13)
3*3*3 =27 = 1 (13)
3*3*3*3 = 81 = 3(13)
3*3*3*3*3 = 243 = 9(13)
3*3*3*3*3*3 = 1 (13)

Do this, and then check that the results are allowed given the constraints you can infer from the order of the Cyclic Group.
 
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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