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Nonabelian dihedral groups and a question in number theory [duplicate]

Is it true that a dihedral group is nonabelian?The center of the dihedral groupOn the centres of the dihedral groupsShowing that Ab(G) need not be completeProve that the dihedral group $D_4$ can not be written as a direct product of two groupsFind all $x$ in $D_4$ such that $ax(a^-1) = b$.Subgroups of generalized dihedral groupsHomomorphisms between Dihedral and Cyclic GroupsFind a permutation that satisfies this congruenceDefining dihedral groups $sigma in S_n: $ something $$Is it only the generator of the group that commutes with all the other elements?List of symmetric group elements from the usual presentation

0

$begingroup$

This question already has an answer here:

• 
  Is it true that a dihedral group is nonabelian?
  
  5 answers
  
  

I'll use a concrete definition of a dihedral group $D_2n$ which emphasizes its group structure:

$D_2n$ consists of distinct elements $r_0,...,r_n-1,s_0,...,s_n-1$ so that for any $i in mathbbZ$ we have $r_i + n = r_i, s_i + n s_i$ and for any $i,j in mathbbZ$ we have $r_ir_j = r_i + j, r_is_j = s_i + j, s_ir_j = s_i - j, s_is_j = r_i - j$.

It's easy to see that all $r_i's$ commute in $D_2n$, hence the only elements for which it is possible not to commute are $s_i$ and $s_j$ or $r_i$ and $s_j$.

$s_is_j = s_js_i$ whenever $i - j equiv j - i bmod n$ and $r_is_j = s_jr_i$ whenever $i + j equiv i - j bmod n$, hence $D_2n$ is noncommutative if and only if either

• there is $0 leq j < n$ so that $nnmid 2j$, or




• there are $0 leq i,j < n$ so that $nnmid 2(i - j)$.

Using this result, it's not hard to establish that $D_2$ and $D_4$ are abelian.

I wonder how we can use this to prove that $D_2n$ is not abelian for $n geq 3$. I'm not interested in a possible geometric proof.

group-theory elementary-number-theory dihedral-groups

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asked Mar 13 at 19:17

Jxt921Jxt921

1,026618

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marked as duplicate by Dietrich Burde group-theory
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Mar 13 at 19:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Is it true that a dihedral group is nonabelian?
  
  5 answers
  
  

I'll use a concrete definition of a dihedral group $D_2n$ which emphasizes its group structure:

$D_2n$ consists of distinct elements $r_0,...,r_n-1,s_0,...,s_n-1$ so that for any $i in mathbbZ$ we have $r_i + n = r_i, s_i + n s_i$ and for any $i,j in mathbbZ$ we have $r_ir_j = r_i + j, r_is_j = s_i + j, s_ir_j = s_i - j, s_is_j = r_i - j$.

It's easy to see that all $r_i's$ commute in $D_2n$, hence the only elements for which it is possible not to commute are $s_i$ and $s_j$ or $r_i$ and $s_j$.

$s_is_j = s_js_i$ whenever $i - j equiv j - i bmod n$ and $r_is_j = s_jr_i$ whenever $i + j equiv i - j bmod n$, hence $D_2n$ is noncommutative if and only if either

• there is $0 leq j < n$ so that $nnmid 2j$, or




• there are $0 leq i,j < n$ so that $nnmid 2(i - j)$.

Using this result, it's not hard to establish that $D_2$ and $D_4$ are abelian.

I wonder how we can use this to prove that $D_2n$ is not abelian for $n geq 3$. I'm not interested in a possible geometric proof.

group-theory elementary-number-theory dihedral-groups

share|cite|improve this question

asked Mar 13 at 19:17

Jxt921Jxt921

1,026618

$endgroup$

marked as duplicate by Dietrich Burde group-theory
Users with the  group-theory badge can single-handedly close group-theory questions as duplicates and reopen them as needed.

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Mar 13 at 19:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Is it true that a dihedral group is nonabelian?
  
  5 answers
  
  

I'll use a concrete definition of a dihedral group $D_2n$ which emphasizes its group structure:

$D_2n$ consists of distinct elements $r_0,...,r_n-1,s_0,...,s_n-1$ so that for any $i in mathbbZ$ we have $r_i + n = r_i, s_i + n s_i$ and for any $i,j in mathbbZ$ we have $r_ir_j = r_i + j, r_is_j = s_i + j, s_ir_j = s_i - j, s_is_j = r_i - j$.

It's easy to see that all $r_i's$ commute in $D_2n$, hence the only elements for which it is possible not to commute are $s_i$ and $s_j$ or $r_i$ and $s_j$.

$s_is_j = s_js_i$ whenever $i - j equiv j - i bmod n$ and $r_is_j = s_jr_i$ whenever $i + j equiv i - j bmod n$, hence $D_2n$ is noncommutative if and only if either

• there is $0 leq j < n$ so that $nnmid 2j$, or




• there are $0 leq i,j < n$ so that $nnmid 2(i - j)$.

Using this result, it's not hard to establish that $D_2$ and $D_4$ are abelian.

I wonder how we can use this to prove that $D_2n$ is not abelian for $n geq 3$. I'm not interested in a possible geometric proof.

group-theory elementary-number-theory dihedral-groups

share|cite|improve this question

asked Mar 13 at 19:17

Jxt921Jxt921

1,026618

$endgroup$

share|cite|improve this question

asked Mar 13 at 19:17

Jxt921Jxt921

1,026618

share|cite|improve this question

asked Mar 13 at 19:17

Jxt921Jxt921

1,026618

asked Mar 13 at 19:17

Jxt921Jxt921

1,026618

asked Mar 13 at 19:17

Jxt921Jxt921

1,026618

2 Answers
2

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oldest

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1

$begingroup$

I bet you can prove that
$$r_1s_1neq s_1r_1$$
It follows easily from your definition.

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answered Mar 13 at 19:35

Matt SamuelMatt Samuel

38.8k63769

$endgroup$

add a comment | 

0

$begingroup$

By definition there is a rotation $r$ in $D_n$, $nge 3$ and a reflection $s$ such that
$$
srs^-1=r^-1neq r.
$$
So $D_n$ is non-abelian. We can also give a second proof by determining the center of $D_n$. A group is abelian iff it equals its center. However, we easily see that $Z(D_n)$ is much smaller than that:

The center of the dihedral group

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answered Mar 13 at 19:44

Dietrich BurdeDietrich Burde

81k648106

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add a comment | 

1

$begingroup$

I bet you can prove that
$$r_1s_1neq s_1r_1$$
It follows easily from your definition.

share|cite|improve this answer

answered Mar 13 at 19:35

Matt SamuelMatt Samuel

38.8k63769

$endgroup$

add a comment | 

1

$begingroup$

I bet you can prove that
$$r_1s_1neq s_1r_1$$
It follows easily from your definition.

share|cite|improve this answer

answered Mar 13 at 19:35

Matt SamuelMatt Samuel

38.8k63769

$endgroup$

add a comment | 

$begingroup$

I bet you can prove that
$$r_1s_1neq s_1r_1$$
It follows easily from your definition.

share|cite|improve this answer

answered Mar 13 at 19:35

Matt SamuelMatt Samuel

38.8k63769

$endgroup$

share|cite|improve this answer

answered Mar 13 at 19:35

Matt SamuelMatt Samuel

38.8k63769

answered Mar 13 at 19:35

Matt SamuelMatt Samuel

38.8k63769

answered Mar 13 at 19:35

Matt SamuelMatt Samuel

38.8k63769

0

$begingroup$

By definition there is a rotation $r$ in $D_n$, $nge 3$ and a reflection $s$ such that
$$
srs^-1=r^-1neq r.
$$
So $D_n$ is non-abelian. We can also give a second proof by determining the center of $D_n$. A group is abelian iff it equals its center. However, we easily see that $Z(D_n)$ is much smaller than that:

The center of the dihedral group

share|cite|improve this answer

answered Mar 13 at 19:44

Dietrich BurdeDietrich Burde

81k648106

$endgroup$

add a comment | 

0

$begingroup$

By definition there is a rotation $r$ in $D_n$, $nge 3$ and a reflection $s$ such that
$$
srs^-1=r^-1neq r.
$$
So $D_n$ is non-abelian. We can also give a second proof by determining the center of $D_n$. A group is abelian iff it equals its center. However, we easily see that $Z(D_n)$ is much smaller than that:

The center of the dihedral group

share|cite|improve this answer

answered Mar 13 at 19:44

Dietrich BurdeDietrich Burde

81k648106

$endgroup$

add a comment | 

$begingroup$

By definition there is a rotation $r$ in $D_n$, $nge 3$ and a reflection $s$ such that
$$
srs^-1=r^-1neq r.
$$
So $D_n$ is non-abelian. We can also give a second proof by determining the center of $D_n$. A group is abelian iff it equals its center. However, we easily see that $Z(D_n)$ is much smaller than that:

The center of the dihedral group

share|cite|improve this answer

answered Mar 13 at 19:44

Dietrich BurdeDietrich Burde

81k648106

$endgroup$

share|cite|improve this answer

answered Mar 13 at 19:44

Dietrich BurdeDietrich Burde

81k648106

answered Mar 13 at 19:44

Dietrich BurdeDietrich Burde

81k648106

answered Mar 13 at 19:44

Dietrich BurdeDietrich Burde

81k648106