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 $n,m \in \mathbb{N}$)?


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 $k$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.