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I am trying to find the number of group homomorphisms from $D_m$ to $\mathbb Z_n$ when $m$ is odd.

Now in MSE I have seen but am unable to find that $Hom(G, A) \equiv Hom(G/[G, G], A)$, where $G/[G,G]$ is referred to as abelianization of $G$, where $A$ is abelian group.

Using this we can say that there is a bijection between $Hom(D_m, \mathbb Z_n)$ and $Hom(D_m/[D_m, D_m], \mathbb Z_n)$. Here $m$ is odd.

But after this I am unable to proceed.

What is $D_m/[D_m, D_m]$ actually ?

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  • $\begingroup$ Do you know a presentation of the dihedral group? $\endgroup$
    – PseudoNeo
    Sep 7, 2015 at 16:26
  • $\begingroup$ :-( No knowledge about it. abelianization I have come to know from MSE only. Nowhere else. $\endgroup$
    – KON3
    Sep 7, 2015 at 16:27
  • $\begingroup$ In this case, I guess that computing the abelianisation will not be easier than computing directly the morphisms $D_m \to \mathbb Z/n$. Hint: the dihedral group is generated by the reflections... $\endgroup$
    – PseudoNeo
    Sep 7, 2015 at 16:37
  • $\begingroup$ @PseudoNeo I have tried that part in this new post math.stackexchange.com/questions/1425554/… would you mind to check it ? $\endgroup$
    – KON3
    Sep 7, 2015 at 16:39

1 Answer 1

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A dihedral group is generated by two reflections $s_1,s_2$. For $D_n$ we have the relation $(s_1s_2)^n=1$. The element $s_1s_2s_1s_2$ is always a commutator, which means that in the abelianization we have $([s_1][s_2])^2=1$ and $([s_1][s_2])^n=1$. If $n$ is odd, this means $[s_1]=[s_2]$ (since the order must divide both $n$ and $2$) and the abelianization is $\mathbb{Z}_2$. If $n$ is even then $[s_1]$ and $[s_2]$ are distinct generators of order $2$ that commute, so the abelianization is $\mathbb{Z}_2\times \mathbb{Z}_2$.

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  • $\begingroup$ I am totally blank now, No idea how to find out Hom$(D_m, \mathbb Z_n)$ :-( $\endgroup$
    – KON3
    Sep 7, 2015 at 17:47
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    $\begingroup$ I answered your other question. Have to go out now so probably won't be available to explain anything else for a few hours. $\endgroup$ Sep 7, 2015 at 17:48
  • $\begingroup$ No problem. Please clear my doubt whenever you will come back. $\endgroup$
    – KON3
    Sep 7, 2015 at 17:52
  • $\begingroup$ I was not aware of this presentation of the dihedral group. That was very helpfull $\endgroup$ Sep 24, 2019 at 20:50
  • $\begingroup$ could I ask a few questions, first by reflections you mean symmetries correct? Also is there an alternate way to show this by showing $[D_n,D_n]$ is generated by $r$ when $n$ odd, and $r^2$ when $n$ even? $\endgroup$
    – homosapien
    Feb 7, 2020 at 22:39

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