What is the abelianization of the dihedral group?












3












$begingroup$


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 ?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know a presentation of the dihedral group?
    $endgroup$
    – PseudoNeo
    Sep 7 '15 at 16:26










  • $begingroup$
    :-( No knowledge about it. abelianization I have come to know from MSE only. Nowhere else.
    $endgroup$
    – Anjan3
    Sep 7 '15 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 '15 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 ?
    $endgroup$
    – Anjan3
    Sep 7 '15 at 16:39
















3












$begingroup$


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 ?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know a presentation of the dihedral group?
    $endgroup$
    – PseudoNeo
    Sep 7 '15 at 16:26










  • $begingroup$
    :-( No knowledge about it. abelianization I have come to know from MSE only. Nowhere else.
    $endgroup$
    – Anjan3
    Sep 7 '15 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 '15 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 ?
    $endgroup$
    – Anjan3
    Sep 7 '15 at 16:39














3












3








3





$begingroup$


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 ?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




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 ?






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 19:22









Matt Samuel

38k63666




38k63666










asked Sep 7 '15 at 16:16









Anjan3Anjan3

2,300922




2,300922












  • $begingroup$
    Do you know a presentation of the dihedral group?
    $endgroup$
    – PseudoNeo
    Sep 7 '15 at 16:26










  • $begingroup$
    :-( No knowledge about it. abelianization I have come to know from MSE only. Nowhere else.
    $endgroup$
    – Anjan3
    Sep 7 '15 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 '15 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 ?
    $endgroup$
    – Anjan3
    Sep 7 '15 at 16:39


















  • $begingroup$
    Do you know a presentation of the dihedral group?
    $endgroup$
    – PseudoNeo
    Sep 7 '15 at 16:26










  • $begingroup$
    :-( No knowledge about it. abelianization I have come to know from MSE only. Nowhere else.
    $endgroup$
    – Anjan3
    Sep 7 '15 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 '15 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 ?
    $endgroup$
    – Anjan3
    Sep 7 '15 at 16:39
















$begingroup$
Do you know a presentation of the dihedral group?
$endgroup$
– PseudoNeo
Sep 7 '15 at 16:26




$begingroup$
Do you know a presentation of the dihedral group?
$endgroup$
– PseudoNeo
Sep 7 '15 at 16:26












$begingroup$
:-( No knowledge about it. abelianization I have come to know from MSE only. Nowhere else.
$endgroup$
– Anjan3
Sep 7 '15 at 16:27




$begingroup$
:-( No knowledge about it. abelianization I have come to know from MSE only. Nowhere else.
$endgroup$
– Anjan3
Sep 7 '15 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 '15 at 16:37




$begingroup$
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 '15 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 ?
$endgroup$
– Anjan3
Sep 7 '15 at 16:39




$begingroup$
@PseudoNeo I have tried that part in this new post math.stackexchange.com/questions/1425554/… would you mind to check it ?
$endgroup$
– Anjan3
Sep 7 '15 at 16:39










1 Answer
1






active

oldest

votes


















5












$begingroup$

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 $mathbb{Z}_2times mathbb{Z}_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am totally blank now, No idea how to find out Hom$(D_m, mathbb Z_n)$ :-(
    $endgroup$
    – Anjan3
    Sep 7 '15 at 17:47






  • 1




    $begingroup$
    I answered your other question. Have to go out now so probably won't be available to explain anything else for a few hours.
    $endgroup$
    – Matt Samuel
    Sep 7 '15 at 17:48










  • $begingroup$
    No problem. Please clear my doubt whenever you will come back.
    $endgroup$
    – Anjan3
    Sep 7 '15 at 17:52











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1 Answer
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1 Answer
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active

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oldest

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active

oldest

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5












$begingroup$

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 $mathbb{Z}_2times mathbb{Z}_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am totally blank now, No idea how to find out Hom$(D_m, mathbb Z_n)$ :-(
    $endgroup$
    – Anjan3
    Sep 7 '15 at 17:47






  • 1




    $begingroup$
    I answered your other question. Have to go out now so probably won't be available to explain anything else for a few hours.
    $endgroup$
    – Matt Samuel
    Sep 7 '15 at 17:48










  • $begingroup$
    No problem. Please clear my doubt whenever you will come back.
    $endgroup$
    – Anjan3
    Sep 7 '15 at 17:52
















5












$begingroup$

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 $mathbb{Z}_2times mathbb{Z}_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am totally blank now, No idea how to find out Hom$(D_m, mathbb Z_n)$ :-(
    $endgroup$
    – Anjan3
    Sep 7 '15 at 17:47






  • 1




    $begingroup$
    I answered your other question. Have to go out now so probably won't be available to explain anything else for a few hours.
    $endgroup$
    – Matt Samuel
    Sep 7 '15 at 17:48










  • $begingroup$
    No problem. Please clear my doubt whenever you will come back.
    $endgroup$
    – Anjan3
    Sep 7 '15 at 17:52














5












5








5





$begingroup$

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 $mathbb{Z}_2times mathbb{Z}_2$.






share|cite|improve this answer









$endgroup$



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 $mathbb{Z}_2times mathbb{Z}_2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 7 '15 at 17:04









Matt SamuelMatt Samuel

38k63666




38k63666












  • $begingroup$
    I am totally blank now, No idea how to find out Hom$(D_m, mathbb Z_n)$ :-(
    $endgroup$
    – Anjan3
    Sep 7 '15 at 17:47






  • 1




    $begingroup$
    I answered your other question. Have to go out now so probably won't be available to explain anything else for a few hours.
    $endgroup$
    – Matt Samuel
    Sep 7 '15 at 17:48










  • $begingroup$
    No problem. Please clear my doubt whenever you will come back.
    $endgroup$
    – Anjan3
    Sep 7 '15 at 17:52


















  • $begingroup$
    I am totally blank now, No idea how to find out Hom$(D_m, mathbb Z_n)$ :-(
    $endgroup$
    – Anjan3
    Sep 7 '15 at 17:47






  • 1




    $begingroup$
    I answered your other question. Have to go out now so probably won't be available to explain anything else for a few hours.
    $endgroup$
    – Matt Samuel
    Sep 7 '15 at 17:48










  • $begingroup$
    No problem. Please clear my doubt whenever you will come back.
    $endgroup$
    – Anjan3
    Sep 7 '15 at 17:52
















$begingroup$
I am totally blank now, No idea how to find out Hom$(D_m, mathbb Z_n)$ :-(
$endgroup$
– Anjan3
Sep 7 '15 at 17:47




$begingroup$
I am totally blank now, No idea how to find out Hom$(D_m, mathbb Z_n)$ :-(
$endgroup$
– Anjan3
Sep 7 '15 at 17:47




1




1




$begingroup$
I answered your other question. Have to go out now so probably won't be available to explain anything else for a few hours.
$endgroup$
– Matt Samuel
Sep 7 '15 at 17:48




$begingroup$
I answered your other question. Have to go out now so probably won't be available to explain anything else for a few hours.
$endgroup$
– Matt Samuel
Sep 7 '15 at 17:48












$begingroup$
No problem. Please clear my doubt whenever you will come back.
$endgroup$
– Anjan3
Sep 7 '15 at 17:52




$begingroup$
No problem. Please clear my doubt whenever you will come back.
$endgroup$
– Anjan3
Sep 7 '15 at 17:52


















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