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Suppose that the number of elements of order $d$ in a group $G$ is denoted by $\eta_d(G)$.

We are willing to find the number of group homomorphisms from $\mathbb Z_m$ to $D_n$ where $D_n$ is dihedral group of order $2n$, $m, n \in \mathbb N$.

We also know that if $C_a$ be a cyclic group of order $a$ then $\#Hom(C_a, G)=\#\{x\in G:x^a=e\}$. Hence by this we can say $$\#Hom(\mathbb Z_{m}, D_n)=\#\{x\in D_n: x^m=e\}=\sum\limits_{d|m}\eta_d(D_n)$$

here is doubt of mine. Note that if we say $x^m=e$ thats means $|x|$ divides $2n$ also. So shall we say the above sum is nothing but $$\sum\limits_{d|(m, 2n)}\eta_d(D_n)?$$

Just answer me if I am wrong. I would like find out the rest of the work. Only I need to know if I am going in the right track.

Thanks in advance

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  • $\begingroup$ Looks good to me. $\eta_d(D_n) = 0$ whenever $d \nmid 2n$. $\endgroup$ – moonlight Sep 7 '15 at 12:15
  • $\begingroup$ Looks good to me too. $x^m=e$ if and only if $|x|$ divides $m$. $\endgroup$ – uniquesolution Sep 7 '15 at 12:16

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