SECTION 6 Cyclic Groups

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SECTION 6 Cyclic Groups

• Recall If G is a group and a?G, then Hann ?Z
  is a subgroup of G. This group is the cyclic
  subgroup ?a? of G generated by a.
• Also, given a group G and an element a in G, if G
  ann ?Z , then a is a generator of G and the
  group G ?a? is cyclic
• Let a be an element of a group G. If the cyclic
  subgroup ?a? is finite, then the order of a is
  the order ?a? of this cyclic subgroup.
  Otherwise, we say that a is of infinite order.

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Elementary Properties of Cyclic Groups

• Theorem
• Every cyclic group is abelian.
• Proof Let G be a cyclic group and let a be a
  generator of G so that
• G ?a? ann ?Z.
• If g1 and g2 are any two elements of G, there
  exists integers r and s such that g1ar and
  g2as. Then
• g1g2 aras ars asr
  asar g2g1,
• So G is abelian.

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Division Algorithm for Z

• Division Algorithm for Z
• If m is a positive integer and n is any integer,
  then there exist unique integers q and r such
  that
• n m q r and
  0?? r lt m
• Here we regard q as the quotient and r as the
  nonnegative remainder when n is divided by m.
• Example
• Find the quotient q and remainder r when 38 is
  divided by 7.
• q5, r3
• Find the quotient q and remainder r when -38 is
  divided by 7.
• q-6, r4

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Theorem

• Theorem
• A subgroup of a cyclic group is cyclic.
• Proof by the division algorithm.
• Corollary
• The subgroups of Z under addition are precisely
  the groups nZ under addition for n?Z.

5
Greatest common divisor

• Let r and s be two positive integers. The
  positive generator d of the cyclic group
• H nr ms n, m ?Z
• Under addition is the greatest common divisor
  (gcd) of r and s. W write d gcd (r, s).
• Note that dnrms for some integers n and m.
  Every integer dividing both r and s divides the
  right-hand side of the equation, and hence must
  be a divisor of d also. Thus d must be the
  largest number dividing both r and s.
• Example Find the gcd of 42 andf 72.
• 6

6
Relatively Prime

• Two positive integers are relatively prime if
  their gcd is 1.
• Fact
• If r and s are relatively prime and if r divides
  sm, then r must divide m.

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The structure of Cyclic Groups

• We can now describe all cyclic groups, up to an
  isomorphism.
• Theorem
• Let G be a cyclic group with generator a. If the
  order of G is infinite, then G is isomorphic to ?
  Z, ?. If G has finite order n, then G is
  isomorphic to ? Zn, n ?.

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Subgroups of Finite Cyclic Groups

• Theorem
• Let G be a cyclic group with n elements and
  generated by a. Let b?G and let bas. Then b
  generates a cyclic subgroup H of G containing n/d
  elements, where d gcd (n, s).
• Also ? as ? ? ar ? if and only gcd (s, n) gcd
  (t, n).
• Example using additive notation, consider in
  Z12, with the generator a1.
• 3 3?1, gcd(3, 12)3, so ? 3 ? has 12/34
  elements. ? 3 ?0, 3, 6, 9
• Furthermore, ? 3 ? ? 9 ? since gcd(3, 12)gcd(9,
  12).
• 8 8?1, gcd (8, 12)4, so ? 8 ? has 12/43
  elements. ? 8 ?0, 4, 8
• 5 5?1, gcd (5, 12)12, so ? 5 ? has 12 elements.
  ? 5 ?Z12.

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Subgroup Diagram of Z18

• Corollary
• If a is a generator of a finite cyclic group G of
  order n, then the other generators of G are the
  elements of the form ar, where r is relatively
  prime to n.
• Example Find all subgroups of Z18 and give their
  subgroup diagram.
• All subgroups are cyclic
• By Corollary, 1 is the generator of Z18, so is 5,
  7, 11, 13, and 17.
• Starting with 2, ? 2 ? 0, 2, 4, 6, 8, 10, 12,
  14, 16 is of order 9, and gcd(2, 18)2gcd(k,
  18) where k is 2, 4, 8, 10, 14, and 16. Thus 2,
  4, 8, 10, 14, and 16 are all generators of ?2?.
• ?3?0, 3, 6, 9, 12, 15 is of order 6, and
  gcd(3, 18)3gcd(k, 18) where k15
• ?6?0, 6, 12 is of order 3, so is 12
• ?9?0, 9 is of order 2

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Subgroup diagram of Z18

• ?1?
• ?2? ?3?
• ?6?
  ?9?
• ?0?