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

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SUMMARY

In the discussion regarding the isomorphism of the automorphism groups of cyclic groups, it is established that Aut(A) for a cyclic group A of order 9 and Aut(B) for a cyclic group B of order 7 cannot be isomorphic due to differing orders. Both Aut(A) and Aut(B) are cyclic groups, but their respective orders are 6 and 6, derived from the formula Aut(Z/nZ) = φ(n), where φ is the Euler's totient function. The confusion arose from a miscalculation regarding the nature of the number 9, which is not prime.

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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.
 

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