Generators and Relations

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Generators and Relations

• Jacob Danton

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Definition

• Let be a group generated by a set
• and be the free
  group on
• . Let be a subset
  of and the smallest normal subgroup
  containing . Then, we say is given by the
  generators and the relations
• if there is an
  isomorphism from onto that carries
  to .

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Notation

• The notation used is
• This is also referred to as the groups
  presentation.

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Theorem 26.3 (Dyck, 1882)

• Let
• 
  
• and
• Then is a homomorphic image of .

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Proof of Dycks Theorem

• Let be the free group on
• be the smallest normal group containing
• and
• the smallest normal group containing
• Note that .

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Proof of Dycks Theorem

• Then and .
• Consider via
• Let .
• Then .
• Thus
• So is well-defined.

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Proof of Dycks Theorem

• Let , then
• Thus, is a homomorphism.
• Let , then
  .
• So, is an epimorphism.
• Thus, is a homomorphic image of
• And, so, a homomorphic image of .

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Corollary to Dycks Theorem Largest Group
Satisfying Defining Relations

• If is a group satisfying the defining
  relations of a finite group and ,
  then is isomorphic to .

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Proof of Corollary

• By Theorem 26.3, we know is a homomorphic
  image of .
• Thus, and, by the supposition we have
  , so .
• Thus the epimorphism is injective and, hence, an
  isomorphism.

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Example 1

• Consider the group, , with presentation
• Then and, thus, .
• So, all we need to do is find a group generated
  by a single element that satisfies the relations
  and has order .

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Example 1

• Clearly, the cyclic group of order works.
• And, thus, .

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Example 2

• Let
• and
• Then
• Consider which is generated by a rotation of
  , , and a horizontal reflection, .
• Notice that ,
• and

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Example 2

• So, satisfies the defining relations of .
  
• Thus, by the previous corollary,

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Example 3

• Consider the group
• Let
• Since , we know and
  .
• Then and, thus, .
• Notice that
• and so
• Thus and

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Example 3

• Notice that by letting and
  ,
• satisfies the defining relations of
  and .
• It may be the largest group to do so, in which
  case .
• In this case, both generators have order 2.
• Thus, if a group of order 8 exists, its
  generators would need to have order 4.

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Example 3

• Consider the group generated by
• and

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Example 3

• So satisfies s defining relations. Also,
  if we write out all the elements generated by
  and , we see that has exactly 8 elements.
• Thus, by the corollary to Dycks Theorem,
• also has 8 elements. This group is known
  as the quaternions and is denoted by .

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Theorem 26.4 Classification of Groups of Order
8

• Up to isomorphism, there are only five groups of
  order 8
• , , , ,
  and .

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Proof of Theorem 26.4

• By the Fundamental Theorem of Finite Abelian
  Groups, we know there are exactly three Abelian
  groups of order 8
• , , and
• So it is only necessary to classify the
  non-Abelian groups.

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Proof of Theorem 26.4

• Suppose is a non-Abelian group of order 8.
• Then 2 or 4.
• Suppose all elements have order 2, then
• Then, would be Abelian.
• So, there must exist an element of order 4 which
  we will call .

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Proof of Theorem 26.4

• Let then, since
  ,
• we have
• Consider
• Since , we know , , ,
  or
• And, since does not commute with ,
• or
• Thus, must be equal to or .

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Proof of Theorem 26.4

• Suppose .
• Since is normal, .
• And, since , or
• However, and, thus,
  is Abelian. So, it must be that
• And, so
• Thus, satisfies the defining relations for
• and .

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Proof of Theorem 26.4

• Suppose , then, using the same argument
  as before, we know .
• Thus,
• And so satisfies the defining relations for
• and .
• So the only non-Abelian groups of order 8 are
  and .

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Groups of Order 12

• By using similar methods, we can classify all
  groups of order 12. They include the two Abelian
  groups, and , in addition to three
  non-Abelian groups
• , , and , the dicyclic group of order
  12. This group has presentation

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Classification of Groups of Order Up to 15

• We can now use this information in addition to
  the Corollary to Theorem 24.2 and Theorem 24.6 to
  classify all groups of order up to 15.
• They are summed up in the following table.

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Classification of Groups of Order Up to 15
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Lemma 1Presentation of Dihedral Groups

• , the dihedral group of order , is
  isomorphic to
  .

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Proof of Lemma 1

• Let , then .
• Thus, since , .
• Consider , where denotes a
  rotation of and a reflection about the
  vertical axis.
• Then
• And, by the corollary to Theorem 26.3,

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Infinite Dihedral Group

• The infinite dihedral group, , is defined as
  .
• The elements can be listed as

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Theorem 26.5Characterization of Dihedral Groups

• Any group generated by a pair of elements of
  order 2 is dihedral.

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Proof of Theorem 26.5

• Let be a group generated by a pair of distinct
  elements, and , where .
• If , then is infinite and satisfies
  the defining relations of .
• Thus, by Theorem 26.3, is the homomorphic
  image of and, thus, is isomorphic to
  for some normal group .

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Proof of Theorem 26.5

• Let be a non-identity element of , then,
  without loss of generality, we have
• or for some non-negative integer .
• If ,
• Thus, since , is
  generated by and , with ,
  , and
• , and so it satisfies the
  defining relations for , a contradiction.

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Proof of Theorem 26.5

• If ,
• Thus, and so
• Since , we again have that
  
• is finite, another
  contradiction.
• Thus, it must be that is the trivial subgroup
  and, therefore, .

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Proof of Theorem 26.5

• Suppose instead that has finite order, say
  .
• Then, we have , , and
• Thus, satisfies the defining relations of
  and, therefore, .

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Homework

• Show that
• Using Example 1 as a guideline, prove