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Suppose $A,B,C$ are finite cyclic groups such that $A = B \times C$, where the orders of $B$ and $C$ are $p$ and $p^2$ respectively, where $p$ is a prime. What are the orders of $End(A)$ and $Aut(A)$?

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Maybe I'm missing something here, but how can $A,B,C$ be cyclic and at the same time $A=B\times C$, $|B|=p$, $|C|=p^2$. The cyclic group of order $p^3$ ($A$) is not isomorphic to the direct product of the cyclic groups of orders $p$ and $p^2$... –  Oliver Braun Dec 9 '12 at 8:26
    
If $A=B\times C$ where $B,C$ have orders $p,p^2$ resp. then $A$ cannot be cyclic. –  anon Dec 9 '12 at 8:26

1 Answer 1

up vote 1 down vote accepted

You sure have a lot of questions:

Hint:

  1. $\text{Hom}(X\times Y,Z)\cong\text{Hom}(X,Z)\times\text{Hom}(Y,Z)$ and $\text{Hom}(X,Y\times Z)\cong\text{Hom}(X,Y)\times\text{Hom}(X,Z)$.

  2. $\text{Hom}(\mathbb{Z}_a,\mathbb{Z}_b)\cong\mathbb{Z}_{(a,b)}$.

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I'm preparing for an exam, and need to have all doubts cleared! Also, what does this notation mean: $\mathbb{Z}_{(a,b)}$ ? –  chubbycantorset Dec 9 '12 at 8:21
    
$(a,b)=\text{gcd}(a,b)$. –  Alex Youcis Dec 9 '12 at 8:24

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