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Unique factorization of dihedral group

Dihedral group of order 8Dihedral group and cyclic group theorem.Dihedral group as a direct productInfinite Dihedral Group.Question about Quaternion group $Q_8$ and Dihedral group $D_8$Dihedral group is supersolvableWhat finite groups have unique cyclic subgroups?Showing that the dihedral group $D_n = {a^i b^j mid 0 <i <6, 0 <j< 2 }$ quotiented by $langle a^k rangle$ is isomorphic to $D_k$Help with understanding proof: every group of order 6 is cclic or dihedral.Compute cosets in the dihedral group

4

1

$begingroup$

My goal is to prove the following about the dihedral group $D_{2n}$:

Prove that every element in $D_{2n}$ has a unique factorization of the form $a^{i}b^{j}$, where $0 leq i < n$ and $j=0$ or $1.$

I know that the cyclic subgroup $left langle a right rangle$ has order $n.$ From this, I know that this has index $2.$ Thus $D_{2n}$ is the disjoint union
$$left langle a right rangle cup left langle a right rangle b.$$

After this, I am pretty stuck. Am I headed in the right direction? What would be the correct way to finish this proof?

Thanks in advance!

abstract-algebra group-theory dihedral-groups

share|cite|improve this question

edited 8 hours ago

Shaun

9,338113684

asked yesterday

MathIsLife12MathIsLife12

613211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you already know that $D_{2n}$ is a disjoint union of $langle arangle$ and $langle arangle b$, then why don't you already know that every element can be written as $a^ib^j$ with $j=0$ or $1$, $0leq ilt n$? And once you know there is at least one such factorization, what is the difficulty in proving uniqueness?
  $endgroup$
  – Arturo Magidin
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, I gather from what you are saying that the proof is almost complete. I just need to show it is unique?
  $endgroup$
  – MathIsLife12
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, you also need to realize why you've already proven the existence of the factorization...
  $endgroup$
  – Arturo Magidin
  yesterday
  

  
  
  
  

add a comment | 

4

1

$begingroup$

My goal is to prove the following about the dihedral group $D_{2n}$:

Prove that every element in $D_{2n}$ has a unique factorization of the form $a^{i}b^{j}$, where $0 leq i < n$ and $j=0$ or $1.$

I know that the cyclic subgroup $left langle a right rangle$ has order $n.$ From this, I know that this has index $2.$ Thus $D_{2n}$ is the disjoint union
$$left langle a right rangle cup left langle a right rangle b.$$

After this, I am pretty stuck. Am I headed in the right direction? What would be the correct way to finish this proof?

Thanks in advance!

abstract-algebra group-theory dihedral-groups

share|cite|improve this question

edited 8 hours ago

Shaun

9,338113684

asked yesterday

MathIsLife12MathIsLife12

613211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you already know that $D_{2n}$ is a disjoint union of $langle arangle$ and $langle arangle b$, then why don't you already know that every element can be written as $a^ib^j$ with $j=0$ or $1$, $0leq ilt n$? And once you know there is at least one such factorization, what is the difficulty in proving uniqueness?
  $endgroup$
  – Arturo Magidin
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, I gather from what you are saying that the proof is almost complete. I just need to show it is unique?
  $endgroup$
  – MathIsLife12
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, you also need to realize why you've already proven the existence of the factorization...
  $endgroup$
  – Arturo Magidin
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

My goal is to prove the following about the dihedral group $D_{2n}$:

Prove that every element in $D_{2n}$ has a unique factorization of the form $a^{i}b^{j}$, where $0 leq i < n$ and $j=0$ or $1.$

I know that the cyclic subgroup $left langle a right rangle$ has order $n.$ From this, I know that this has index $2.$ Thus $D_{2n}$ is the disjoint union
$$left langle a right rangle cup left langle a right rangle b.$$

After this, I am pretty stuck. Am I headed in the right direction? What would be the correct way to finish this proof?

Thanks in advance!

abstract-algebra group-theory dihedral-groups

share|cite|improve this question

edited 8 hours ago

Shaun

9,338113684

asked yesterday

MathIsLife12MathIsLife12

613211

$endgroup$

share|cite|improve this question

edited 8 hours ago

Shaun

9,338113684

asked yesterday

MathIsLife12MathIsLife12

613211

share|cite|improve this question

edited 8 hours ago

Shaun

9,338113684

asked yesterday

MathIsLife12MathIsLife12

613211

edited 8 hours ago

Shaun

9,338113684

edited 8 hours ago

Shaun

9,338113684

asked yesterday

MathIsLife12MathIsLife12

613211

asked yesterday

MathIsLife12MathIsLife12

613211

1 Answer
1

active

oldest

votes

5

$begingroup$

You want products of the form:

$a^{i}b^{j}$, where $0 leq i < n$ and $j=0$ or $1.$

there are $2n$ of these: $a^0,a^1,ldots,a^{n-1}$ and $a^0b,a^1b,ldots,a^{n-1}b$. These $2n$ products each give you an element of $D_{2n}$ so as long as they all represent distinct elements, the factorization is unique.

The $n$ products of the form $a^i$ are distinct because $|a| = n$. Likewise, the elements $a^ib$ are distinct because $a^{i_1}b = a^{i_2}b$ implies $a^{i_1} = a^{i_2}$ by cancellation, and then $i_1 = i_2$ since $0leq i leq n$. Finally, there's no $a^{i_1} b = a^{i_2}$ because that would imply that $b in left<aright>$.

That means your $2n$ products are in 1-to-1 correspondence with the elements of $D_{2n}$; each element of $D_{2n}$ has a unique representative as one of the products.

share|cite|improve this answer

answered yesterday

orlandpmorlandpm

4,70921838

$endgroup$

add a comment | 

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5

$begingroup$

You want products of the form:

$a^{i}b^{j}$, where $0 leq i < n$ and $j=0$ or $1.$

there are $2n$ of these: $a^0,a^1,ldots,a^{n-1}$ and $a^0b,a^1b,ldots,a^{n-1}b$. These $2n$ products each give you an element of $D_{2n}$ so as long as they all represent distinct elements, the factorization is unique.

The $n$ products of the form $a^i$ are distinct because $|a| = n$. Likewise, the elements $a^ib$ are distinct because $a^{i_1}b = a^{i_2}b$ implies $a^{i_1} = a^{i_2}$ by cancellation, and then $i_1 = i_2$ since $0leq i leq n$. Finally, there's no $a^{i_1} b = a^{i_2}$ because that would imply that $b in left<aright>$.

That means your $2n$ products are in 1-to-1 correspondence with the elements of $D_{2n}$; each element of $D_{2n}$ has a unique representative as one of the products.

share|cite|improve this answer

answered yesterday

orlandpmorlandpm

4,70921838

$endgroup$

add a comment | 

5

$begingroup$

You want products of the form:

$a^{i}b^{j}$, where $0 leq i < n$ and $j=0$ or $1.$

there are $2n$ of these: $a^0,a^1,ldots,a^{n-1}$ and $a^0b,a^1b,ldots,a^{n-1}b$. These $2n$ products each give you an element of $D_{2n}$ so as long as they all represent distinct elements, the factorization is unique.

The $n$ products of the form $a^i$ are distinct because $|a| = n$. Likewise, the elements $a^ib$ are distinct because $a^{i_1}b = a^{i_2}b$ implies $a^{i_1} = a^{i_2}$ by cancellation, and then $i_1 = i_2$ since $0leq i leq n$. Finally, there's no $a^{i_1} b = a^{i_2}$ because that would imply that $b in left<aright>$.

That means your $2n$ products are in 1-to-1 correspondence with the elements of $D_{2n}$; each element of $D_{2n}$ has a unique representative as one of the products.

share|cite|improve this answer

answered yesterday

orlandpmorlandpm

4,70921838

$endgroup$

add a comment | 

$begingroup$

You want products of the form:

$a^{i}b^{j}$, where $0 leq i < n$ and $j=0$ or $1.$

there are $2n$ of these: $a^0,a^1,ldots,a^{n-1}$ and $a^0b,a^1b,ldots,a^{n-1}b$. These $2n$ products each give you an element of $D_{2n}$ so as long as they all represent distinct elements, the factorization is unique.

The $n$ products of the form $a^i$ are distinct because $|a| = n$. Likewise, the elements $a^ib$ are distinct because $a^{i_1}b = a^{i_2}b$ implies $a^{i_1} = a^{i_2}$ by cancellation, and then $i_1 = i_2$ since $0leq i leq n$. Finally, there's no $a^{i_1} b = a^{i_2}$ because that would imply that $b in left<aright>$.

That means your $2n$ products are in 1-to-1 correspondence with the elements of $D_{2n}$; each element of $D_{2n}$ has a unique representative as one of the products.

share|cite|improve this answer

answered yesterday

orlandpmorlandpm

4,70921838

$endgroup$

share|cite|improve this answer

answered yesterday

orlandpmorlandpm

4,70921838

answered yesterday

orlandpmorlandpm

4,70921838

answered yesterday

orlandpmorlandpm

4,70921838