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A question about dihedral group

Prove a group generated by two involutions is dihedralInfinite Dihedral Group.How to show that this group presentation is isomorphic to dihedral group?Trying to understand group presentations using the example of the Dihedral groupObtaining a presentation of the dihedral group from a semidirect productDihedral Group question involving $D_{4n}$A question about isomorphic of dihedral groupHow to prove isomorphism with the Dihedral group

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2

1

$begingroup$

The presentation of the dihedral group is
$$
D_{2n}= langle r,s mid r^{n}=s^{2}=1, (sr)^2=1 rangle.
$$
Now, let $G$ be a group of order $2n$. Is it true that if $G$ contains two elements $a,b$ such that $ord(a)=n$, $ord(b)=2$ and $ord(ba)=2$, then $Gcong D_{2n}$ ?

Edit: If not, what are the conditions that $G$ has to fulfill in order to be isomorphic to $D_{2n}$ ?

group-theory group-presentation dihedral-groups finitely-generated

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edited 5 hours ago

Shaun

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asked 9 hours ago

boazboaz

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$endgroup$

add a comment | 

2

1

$begingroup$

The presentation of the dihedral group is
$$
D_{2n}= langle r,s mid r^{n}=s^{2}=1, (sr)^2=1 rangle.
$$
Now, let $G$ be a group of order $2n$. Is it true that if $G$ contains two elements $a,b$ such that $ord(a)=n$, $ord(b)=2$ and $ord(ba)=2$, then $Gcong D_{2n}$ ?

Edit: If not, what are the conditions that $G$ has to fulfill in order to be isomorphic to $D_{2n}$ ?

group-theory group-presentation dihedral-groups finitely-generated

share|cite|improve this question

edited 5 hours ago

Shaun

11.6k1212 gold badges3737 silver badges9292 bronze badges

asked 9 hours ago

boazboaz

2,45088 silver badges1414 bronze badges

$endgroup$

add a comment | 

$begingroup$

The presentation of the dihedral group is
$$
D_{2n}= langle r,s mid r^{n}=s^{2}=1, (sr)^2=1 rangle.
$$
Now, let $G$ be a group of order $2n$. Is it true that if $G$ contains two elements $a,b$ such that $ord(a)=n$, $ord(b)=2$ and $ord(ba)=2$, then $Gcong D_{2n}$ ?

Edit: If not, what are the conditions that $G$ has to fulfill in order to be isomorphic to $D_{2n}$ ?

group-theory group-presentation dihedral-groups finitely-generated

share|cite|improve this question

edited 5 hours ago

Shaun

11.6k1212 gold badges3737 silver badges9292 bronze badges

asked 9 hours ago

boazboaz

2,45088 silver badges1414 bronze badges

$endgroup$

share|cite|improve this question

edited 5 hours ago

Shaun

11.6k1212 gold badges3737 silver badges9292 bronze badges

asked 9 hours ago

boazboaz

2,45088 silver badges1414 bronze badges

share|cite|improve this question

edited 5 hours ago

Shaun

11.6k1212 gold badges3737 silver badges9292 bronze badges

asked 9 hours ago

boazboaz

2,45088 silver badges1414 bronze badges

edited 5 hours ago

Shaun

11.6k1212 gold badges3737 silver badges9292 bronze badges

edited 5 hours ago

Shaun

11.6k1212 gold badges3737 silver badges9292 bronze badges

asked 9 hours ago

boazboaz

2,45088 silver badges1414 bronze badges

asked 9 hours ago

boazboaz

2,45088 silver badges1414 bronze badges

2 Answers
2

active

oldest

votes

4

$begingroup$

YES, it is. Let's call $G_0$ your presentation of $D_{2n}$. The universal property of $G_0$ tells you that there is a morphism $G_0to G$ which sends $r$ to $a$ dand $s$ to $b$.

Claim. $G$ is generated by $a$ and $b$.

For, it is enough to show that the various elements $a^k, ba^ell$ are pairwise distinct (where $0leq k,ellleq n-1$).

The only non trivial thing to prove is that $a^k= ba^ell $ for some $k,ell$ cannot happen. Assume the contrary, so that $b=a^m, m=k-ell$, so $-(n-1)leq mleq n-1$.
Taking inverse and using the fact that $b$ has order $2$, one may assume that $0leq mleq n-1$.

The order of $b$ is $2$, so if $n$ is odd it cannot happen. Write $n=2r$. Then $b=a^r$, the unique element of order of the cyclic group generated by $a$. Now $ba=a^{r+1}$, which has order $2r/gcd(2r,r+1)$. Now $gcd (2r,r+1)$ is $1$ or $2$ since $2(r+1)-2r=2$. Consequently, $o(ba)=2r$ or
$r$.

Therefore, if $rgeq 3$ you have a contradiction. If $r=2$, o$(a^3)=4$ and you have a contradiction again.

To sum up, $G$ is generated by $a$ and $b$.
Hence the morphism above is surjective, hence bijective.

share|cite|improve this answer

answered 9 hours ago

GreginGreGreginGre

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$endgroup$

add a comment | 

3

$begingroup$

The answer is yes

Proof:

Note that $[G:langle arangle]=2$ so $langle a rangle$ is normal in $G$.

We also have $baba=1$ so $b^{-1}ab=a^{-1}$.

This means $Gcong C_nrtimes C_2=D_{2n}$

share|cite|improve this answer

answered 9 hours ago

Robert ChamberlainRobert Chamberlain

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$endgroup$

add a comment | 

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4

$begingroup$

YES, it is. Let's call $G_0$ your presentation of $D_{2n}$. The universal property of $G_0$ tells you that there is a morphism $G_0to G$ which sends $r$ to $a$ dand $s$ to $b$.

Claim. $G$ is generated by $a$ and $b$.

For, it is enough to show that the various elements $a^k, ba^ell$ are pairwise distinct (where $0leq k,ellleq n-1$).

The only non trivial thing to prove is that $a^k= ba^ell $ for some $k,ell$ cannot happen. Assume the contrary, so that $b=a^m, m=k-ell$, so $-(n-1)leq mleq n-1$.
Taking inverse and using the fact that $b$ has order $2$, one may assume that $0leq mleq n-1$.

The order of $b$ is $2$, so if $n$ is odd it cannot happen. Write $n=2r$. Then $b=a^r$, the unique element of order of the cyclic group generated by $a$. Now $ba=a^{r+1}$, which has order $2r/gcd(2r,r+1)$. Now $gcd (2r,r+1)$ is $1$ or $2$ since $2(r+1)-2r=2$. Consequently, $o(ba)=2r$ or
$r$.

Therefore, if $rgeq 3$ you have a contradiction. If $r=2$, o$(a^3)=4$ and you have a contradiction again.

To sum up, $G$ is generated by $a$ and $b$.
Hence the morphism above is surjective, hence bijective.

share|cite|improve this answer

answered 9 hours ago

GreginGreGreginGre

1,77233 silver badges1010 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

YES, it is. Let's call $G_0$ your presentation of $D_{2n}$. The universal property of $G_0$ tells you that there is a morphism $G_0to G$ which sends $r$ to $a$ dand $s$ to $b$.

Claim. $G$ is generated by $a$ and $b$.

For, it is enough to show that the various elements $a^k, ba^ell$ are pairwise distinct (where $0leq k,ellleq n-1$).

The only non trivial thing to prove is that $a^k= ba^ell $ for some $k,ell$ cannot happen. Assume the contrary, so that $b=a^m, m=k-ell$, so $-(n-1)leq mleq n-1$.
Taking inverse and using the fact that $b$ has order $2$, one may assume that $0leq mleq n-1$.

The order of $b$ is $2$, so if $n$ is odd it cannot happen. Write $n=2r$. Then $b=a^r$, the unique element of order of the cyclic group generated by $a$. Now $ba=a^{r+1}$, which has order $2r/gcd(2r,r+1)$. Now $gcd (2r,r+1)$ is $1$ or $2$ since $2(r+1)-2r=2$. Consequently, $o(ba)=2r$ or
$r$.

Therefore, if $rgeq 3$ you have a contradiction. If $r=2$, o$(a^3)=4$ and you have a contradiction again.

To sum up, $G$ is generated by $a$ and $b$.
Hence the morphism above is surjective, hence bijective.

share|cite|improve this answer

answered 9 hours ago

GreginGreGreginGre

1,77233 silver badges1010 bronze badges

$endgroup$

add a comment | 

$begingroup$

YES, it is. Let's call $G_0$ your presentation of $D_{2n}$. The universal property of $G_0$ tells you that there is a morphism $G_0to G$ which sends $r$ to $a$ dand $s$ to $b$.

Claim. $G$ is generated by $a$ and $b$.

For, it is enough to show that the various elements $a^k, ba^ell$ are pairwise distinct (where $0leq k,ellleq n-1$).

The only non trivial thing to prove is that $a^k= ba^ell $ for some $k,ell$ cannot happen. Assume the contrary, so that $b=a^m, m=k-ell$, so $-(n-1)leq mleq n-1$.
Taking inverse and using the fact that $b$ has order $2$, one may assume that $0leq mleq n-1$.

The order of $b$ is $2$, so if $n$ is odd it cannot happen. Write $n=2r$. Then $b=a^r$, the unique element of order of the cyclic group generated by $a$. Now $ba=a^{r+1}$, which has order $2r/gcd(2r,r+1)$. Now $gcd (2r,r+1)$ is $1$ or $2$ since $2(r+1)-2r=2$. Consequently, $o(ba)=2r$ or
$r$.

Therefore, if $rgeq 3$ you have a contradiction. If $r=2$, o$(a^3)=4$ and you have a contradiction again.

To sum up, $G$ is generated by $a$ and $b$.
Hence the morphism above is surjective, hence bijective.

share|cite|improve this answer

answered 9 hours ago

GreginGreGreginGre

1,77233 silver badges1010 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

GreginGreGreginGre

1,77233 silver badges1010 bronze badges

answered 9 hours ago

GreginGreGreginGre

1,77233 silver badges1010 bronze badges

answered 9 hours ago

GreginGreGreginGre

1,77233 silver badges1010 bronze badges

3

$begingroup$

The answer is yes

Proof:

Note that $[G:langle arangle]=2$ so $langle a rangle$ is normal in $G$.

We also have $baba=1$ so $b^{-1}ab=a^{-1}$.

This means $Gcong C_nrtimes C_2=D_{2n}$

share|cite|improve this answer

answered 9 hours ago

Robert ChamberlainRobert Chamberlain

5,01911 gold badge77 silver badges2121 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

The answer is yes

Proof:

Note that $[G:langle arangle]=2$ so $langle a rangle$ is normal in $G$.

We also have $baba=1$ so $b^{-1}ab=a^{-1}$.

This means $Gcong C_nrtimes C_2=D_{2n}$

share|cite|improve this answer

answered 9 hours ago

Robert ChamberlainRobert Chamberlain

5,01911 gold badge77 silver badges2121 bronze badges

$endgroup$

add a comment | 

$begingroup$

The answer is yes

Proof:

Note that $[G:langle arangle]=2$ so $langle a rangle$ is normal in $G$.

We also have $baba=1$ so $b^{-1}ab=a^{-1}$.

This means $Gcong C_nrtimes C_2=D_{2n}$

share|cite|improve this answer

answered 9 hours ago

Robert ChamberlainRobert Chamberlain

5,01911 gold badge77 silver badges2121 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Robert ChamberlainRobert Chamberlain

5,01911 gold badge77 silver badges2121 bronze badges

answered 9 hours ago

Robert ChamberlainRobert Chamberlain

5,01911 gold badge77 silver badges2121 bronze badges

answered 9 hours ago

Robert ChamberlainRobert Chamberlain

5,01911 gold badge77 silver badges2121 bronze badges