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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 $p^2$...2012-12-09
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    If $A=B\times C$ where $B,C$ have orders $p,p^2$ resp. then $A$ cannot be cyclic.2012-12-09

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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: $\mathbb{Z}_{(a,b)}$ ?2012-12-09
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    $(a,b)=\text{gcd}(a,b)$.2012-12-09
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    @Alex Youcis: Do you have a proof-reference for "2." above?2017-12-28
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    @Alex Youcis: Do you have a proof-reference for "1." above?2017-12-28
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    @Alex Youcis: Are you sure that $\text{Hom}(X\times Y,Z)\equiv \text{Hom}(X,Z)\times \text{Hom}(Y,Z)$ holds also for non-abelian $Z$?2017-12-29