This is the situation:
Suppose we have a finitely generated group $G= \mathbb{Z}^r \times E$ with $E$ it's tortion subgroup and $m=$ exponent of $E$ i.e. the least natural number such that $x^m=1$ for each $x\in E$. Suppose you want to enquire on the set $\#Hom(G, C_k)$ where $C_k$ is a finite ciclic group of order $k$. Is there anything we can say about $\#Hom(G, C_k)$ with respect to the values of $m$ and $k$? Maybe something like "if they are coprime then....".
Also, a homomorphism between cyclic groups must send generator to generator. In this case, where G is finitely generated $<a_1, \dots a_k>$ and $C_k= <c>$, a group homomorphism must send $a_i \mapsto c$ (right?) so can I say that we have at leas $k$ morphisms?
Thank you all!

  • $\begingroup$ A homomorphism need not send a generator to a generator. If $m$ and $k$ are coprime, then all of $E$ must be sent to $1$, so you are looking at the number of homomorphisms from $\mathbb{Z}^r$, which is just $k^r$. $\endgroup$ – Tobias Kildetoft Jun 27 '15 at 16:26
  • $\begingroup$ @Tobias Kildetoft Thant is a good result! Could you explaint, briefly, why it would work this way? $\endgroup$ – Tyche Jun 27 '15 at 16:34
  • $\begingroup$ If $x$ has finite order $m$ then the image of $x$ under a homomorphism must have order dividing $m$, which in the coprime case means that the image must have order $1$. The second part is the universal property of the free abelian group on $r$ generators. $\endgroup$ – Tobias Kildetoft Jun 27 '15 at 16:37

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