I am willing to find out the number of group homomorphisms from $D_{12}$ to $D_{18}$ where $D_m:=\langle r_m, f_m: r_m^m=f_m^2=(r_mf_m)^2=e_m \rangle$ is the standrard dihedral group of order $2m$.

Basically I am motivated from this paper. I have studied and understood the first two cases. But for the third case, I am not getting any clue. SO this is why I have started numerical example.

Let $\rho:D_{12}\rightarrow D_{18}$ be a group homomoprhism.

So we can say that $\#$Hom$(D_{12}, D_{18})=$ trivial + non-trivial group homomoprhisms. Here non-trivial group homomorphisms = number of possible choice for $\rho(f_m)\times$ number of possible choices for $\rho(r_m)$.

here according to rule it is supposed to be $4+4\times 18+18\times (12,18)$.

BUt I want to understand the answer step by step. So I started from the begining.

Here $\rho(r_{12})$ has the possibilities $e_{18}, r_{18}^\alpha$ with $1\leq \alpha\leq 17$. BUt I dont understand why in the paper it is said another possibilities $r_{18}^\beta f_{18}$.

How to count the homomorphisms ? Please enlighten me. thanks in advance

My Attempt: Hom$(D_{12}, D_{18})=Hom(D_{12}/[D_{12}, D_{12}], D_{18})=Hom(\mathbb Z_2, D_{18})$. And we know that $\#Hom(\mathbb Z_n, G)=\#\{x\in G: x^n = e_G\}$.

SO if we use this theorem, we then can conclude that required number of nothing but $\#\{x\in D_{18}: x^2=e\}$.

Is it correct ? Not sure if this will give me correct answer according to the formula provided in the paper stated above

  • $\begingroup$ So does abelianization will save the day or not , please let me know. $\endgroup$
    – KON3
    Sep 4, 2015 at 16:23
  • $\begingroup$ Between , the paper I mentioned for some reason is notg opening. Here is the link arxiv.org/pdf/1201.2363.pdf $\endgroup$
    – KON3
    Sep 4, 2015 at 16:33
  • $\begingroup$ From first principles it's probably easiest to use Coxeter generators (at least using them I see a way to definitely be able to compute the number of homomorphisms). $\endgroup$ Sep 4, 2015 at 16:57
  • $\begingroup$ Would you mind to give some link on "coxeter generators" please? have no idea about it. never heard of it either. Please provide some more information $\endgroup$
    – KON3
    Sep 4, 2015 at 17:00
  • 1
    $\begingroup$ You say that $\hom(D_{12},D_{18})=\hom(D_{12}/[D_{12},D_{12}],D_{18})$? Firstly "$=$" should be replaced by $\cong$. Assuming that you are saying that precomposing the homomorphisms in the second $\hom$ set with the canonical quotient $D_{12} \to D_{12}/[D_{12},D_{12}]$ is an isomorphism this would mean that the image of every homomorphism from $D_{12}$ to $D_{18}$ is abelian. Let $\rho(r_{12})=r_{18}^3$ and $\rho(f_{12})=f_18$, then does this determine a homomorphism? If so does this homomorphism have abelian image? $\endgroup$
    – Nex
    Oct 29, 2015 at 7:41


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