Is Aut(A) Isomorphic to Aut(B) for Cyclic Groups of Different Orders?

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Homework Statement


If A=<x> is a cyclic group of order 9 and B=<y> is a cyclic group of order 7. Deduce that Aut(A) is isomorphic to Aut(B)


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The Attempt at a Solution


I already proved that Aut(A) and Aut(B) are cyclic but I don't understand how they can be isomorphic if they don't have the same order. Also the groups are don't have the same amount of generators since both groups are cyclic so every element of both Aut(A) and Aut(B) are generators.
 
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chibulls59 said:
I already proved that Aut(A) and Aut(B) are cyclic but I don't understand how they can be isomorphic if they don't have the same order.
They can't be. So you've either shown the problem is in error, of you've computed Aut(A) and Aut(B) incorrectly.
 
Oh I got it, for some reason I thought 9 was a prime number.
 
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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