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I was asked to describe the group of the endomorphism of $G=C_{p} \times C_{p^2}$, with p prime ($C_n$ is the cyclic group of order $n$).

I started setting (g,1) and (1,h) as generators of the subgroups $(C_{p},1)$ and $ (1,C_{p^2})$ so every endomorphism $\varphi$ is determined by it's value on these elements. $$\varphi((g,1))=g^ah^b$$ $$\varphi((1,g))=g^ch^d$$ so I count $p*p^2*p*p^2$ different results and so there are $p^6$ different endomorphisms. Is this correct? How can I say something about the algebraic structure of $\mathrm{End}(G)$?

This is the first exercise of this type I have to deal with, so I don't know the general strategies to solve it efficiently. Can someone please explain me how to work with endomorphism groups in this specific case (or in the slightly generalization of $C_n\times C_m$ with $n,m \in \mathbb{N}$)?

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There is a restriction: the element $g$ can only be mapped onto an element of period a divisor of $p$. There are $p^{2}$ of these in $G$. The element $h$ must be mapped onto an element of period a divisor of $p^{2}$, but all elements of $G$ satisfy this. So the total number of endomorphisms is $p^{2} \cdot p^{3} = p^{5}$.

Explicitly, an endomorphism $\varphi$ maps $$ g \mapsto g^{a} h^{p b}, \qquad h \mapsto g^{c} h^{d}, $$ where $0 \le d < p^{2}$ and $0 \le a, b, c < p$.

In the general case, again you have only to make sure that you are mapping an generator of order $k$ onto an element of order dividing $k$.

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  • $\begingroup$ Thank you very much, is there a way to describe its structure as a group starting from the structure of the group G? (in analogy with the theorems regarding the structure of Aut(G) starting from G?) $\endgroup$ – Riccardo Sep 5 '13 at 16:46
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    $\begingroup$ @RicPed, this is described for instance in Jacobson's Basic Algebra I, but see for instance math.mcgill.ca/labute/courses/371.98/end.pdf or other online resources. $\endgroup$ – Andreas Caranti Sep 5 '13 at 21:23
  • $\begingroup$ The pdf is exactly what i need, thank you again $\endgroup$ – Riccardo Sep 5 '13 at 23:03

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