How do homomorphisms from C_6 to Aut(C_n) work for n=12 and n=16?

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Describe explicitly all homomorphisms

h: C_6 ----> Aut(C_n)

The question asks when n=12,16

I was wondering if someone could explain how to do this? I've looked through the notes but struggling a tad

I think I could do this if it said for instance h: C_6 ----> C_n but Aut(C_12) = C_2 x C_2 which is a little confusing

Thanks
 
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Can you articulate why you think finding homomorphisms
C_6 ---> C_2 x C_2​
is hard and confusing, but finding homomorphisms
C_6 ---> C_n​
is not?
 
Hurkyl said:
Can you articulate why you think finding homomorphisms
C_6 ---> C_2 x C_2​
is hard and confusing, but finding homomorphisms
C_6 ---> C_n​
is not?

Lack of understanding & having an example for the latter but not the former xD

I think I've figured it out though

for n = 12 there are 4 homomorphisms

h_1(x) = 1
h_2(x) = z
h_3(x) = y
h_4(x) = zy

for n = 16 there are 8 homomorphisms

h_1(x) = 1
h_2(x) = z
h_3(x) = z^2
h_4(x) = z^3
h_5(x) = y
h_6(x) = zy
h_7(x) = z^2y
h_8(x) = z^3y

Would that be correct?

For the next question all homomorphism h: C_5 --> Aut(C_11) I get 5

h_1(x) = 1
h_2(x) = z^2
h_3(x) = z^4
h_4(x) = z^6
h_5(x) = z^8
 
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Can you say why you think that's the answer? Why must any homomorphism be named on that list, and why is everything named on that list a homomorphism?
 
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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