Bidirected edge-maximality of power graphs of finite cyclic groups

Transcription

1 Bidirected edge-maximality of power graphs of finite cyclic groups Brian Curtin 1 Gholam Reza Pourgholi 2 1 Department of Mathematics and Statistics University of South Florida 2 School of Mathematics, Statistics and Computer Science University of Tehran Modern Trends in Algebraic Graph Theory Villanova University June 5th, 2014

2 Power Graphs G : finite group of order n

3 Power Graphs G : finite group of order n P (G) : directed power graph V (G) = G E (G) = {(g, h) g, h G, h g {g}}

4 Power Graphs G : finite group of order n P (G) : directed power graph V (G) = G E (G) = {(g, h) g, h G, h g {g}} bidirected edges E (G) = {{g, h} (g, h) E (G) and (h, g) E (G)}

5 Power Graphs G : finite group of order n P (G) : directed power graph V (G) = G E (G) = {(g, h) g, h G, h g {g}} bidirected edges E (G) = {{g, h} (g, h) E (G) and (h, g) E (G)} Lemma bidirected edge {g, h} iff g h generate the same subgroup.

6 Power Graphs P (Z6 ) 0

7 Power Graphs P (Z6 ) 3 0

8 Power Graphs P (Z6 ) 3 0

9 Power Graphs P (Z6 ) 2 3 0

10 Power Graphs P (Z6 )

11 Power Graphs P (Z6 )

12 Power Graphs P (Z6 )

13 Power Graphs P (Z6 )

14 Power Graphs P (Z6 )

15 Power Graphs P (Z6 )

16 Power Graphs P (Z6 )

17 Power Graphs for groups of order 8 P (C8 )

18 Power Graphs for groups of order 8 P (C8 ) P (C4 C 2 )

19 Power Graphs for groups of order 8 P (C8 ) P (C4 C 2 ) P (C2 C 2 C 2 )

20 Power Graphs for groups of order 8 P (C8 ) P (C4 C 2 ) P (C2 C 2 C 2 ) P (Q) i j -j 1-1 -i -k k

21 Power Graphs for groups of order 8 P (C8 ) P (C4 C 2 ) P (C2 C 2 C 2 ) P (Q) P (D8 ) i j -j ϕρ 2 ϕρ ϕρ e ϕ -i -k k ρ 2 ρ 3 ρ

22 Bidirectional edges Notation o(g) : order of g φ : Euler totient

23 Bidirectional edges Notation o(g) : order of g φ : Euler totient Count bidirected edges

24 Bidirectional edges Notation o(g) : order of g φ : Euler totient Count bidirected edges g : cyclic, order o(g),

25 Bidirectional edges Notation o(g) : order of g φ : Euler totient Count bidirected edges g : cyclic, order o(g), φ(o(g)) distinct generators

26 Bidirectional edges Notation o(g) : order of g φ : Euler totient Count bidirected edges g : cyclic, order o(g), φ(o(g)) distinct generators g is in φ(o(g)) 1 bidirected edges

27 Bidirectional edges Notation o(g) : order of g φ : Euler totient Count bidirected edges g : cyclic, order o(g), φ(o(g)) distinct generators g is in φ(o(g)) 1 bidirected edges Summing over G double counts

28 Bidirectional edges Notation o(g) : order of g φ : Euler totient Count bidirected edges g : cyclic, order o(g), φ(o(g)) distinct generators g is in φ(o(g)) 1 bidirected edges Summing over G double counts Lemma E (G) = 1 (φ(o(g)) 1) (1) 2 g G

29 A group sum Definition φ(g) = g G φ(o(g)). (2)

30 A group sum Definition φ(g) = g G φ(o(g)). (2) Corollary E (G) = φ(g) G 2 (3)

31 A group sum Definition φ(g) = g G φ(o(g)). (2) Corollary E (G) = φ(g) G 2 (3) Notation C n : cyclic group of order n Compare φ(g), φ(c n )

32 Results Main Theorem (BC, GR Pourgholi) Among finite groups of given order, the cyclic group has the maximum number of bidirectional edges in its directed power graph.

33 Results Main Theorem (BC, GR Pourgholi) Among finite groups of given order, the cyclic group has the maximum number of bidirectional edges in its directed power graph. Theorem (H Amiri, SM Jafarian Amiri, IM Isaacs) (Lindsey) Among finite groups of given order, the cyclic group has the maximum number of edges in its directed power graph.

34 Results Main Theorem (BC, GR Pourgholi) Among finite groups of given order, the cyclic group has the maximum number of bidirectional edges in its directed power graph. Theorem (H Amiri, SM Jafarian Amiri, IM Isaacs) (Lindsey) Among finite groups of given order, the cyclic group has the maximum number of edges in its directed power graph. Theorem (BC, GR Pourgholi) Among finite groups of given order, the cyclic group has the maximum number of edges in its undirected power graph.

35 Results restated Main Theorem (BC, GR Pourgholi) φ(g) φ(c n )

36 Results restated Main Theorem (BC, GR Pourgholi) i.e. φ(g) φ(c n ) g G φ(o(g)) z C n φ(o(z))

37 Results restated Main Theorem (BC, GR Pourgholi) i.e. φ(g) φ(c n ) g G φ(o(g)) z C n φ(o(z)) equality iff G = C n

38 Results restated Main Theorem (BC, GR Pourgholi) i.e. φ(g) φ(c n ) g G φ(o(g)) z C n φ(o(z)) equality iff G = C n Theorem (H Amiri, SM Jafarian Amiri, IM Isaacs) (Lindsey) g G o(g) z C n o(z) equality iff G = C n

39 Results restated Main Theorem (BC, GR Pourgholi) i.e. φ(g) φ(c n ) g G φ(o(g)) z C n φ(o(z)) equality iff G = C n Theorem (H Amiri, SM Jafarian Amiri, IM Isaacs) (Lindsey) g G o(g) z C n o(z) equality iff G = C n Theorem (BC, GR Pourgholi) g G 2o(g) φ(o(g)) z C n 2o(z) φ(o(z)), equality iff G = C n

40 φ(c n ) Notation n = p α 1 1 pα 2 2 pα k k p 1 < p 2 < < p k primes α 1, α 2,..., α k Z +

41 φ(c n ) Notation n = p α 1 1 pα 2 2 pα k k p 1 < p 2 < < p k primes α 1, α 2,..., α k Z + Lemma φ(c n ) = d n φ(d)2 = k h=1 p 2α h h (p h 1)+2 p h +1

42 φ(c n ) Notation n = p α 1 1 pα 2 2 pα k k p 1 < p 2 < < p k primes α 1, α 2,..., α k Z + Lemma φ(c n ) = d n φ(d)2 = k h=1 p 2α h h (p h 1)+2 p h +1 Definition Q = k p h +1 h=1 p h 1

43 φ(c n ) Notation n = p α 1 1 pα 2 2 pα k k p 1 < p 2 < < p k primes α 1, α 2,..., α k Z + Lemma φ(c n ) = d n φ(d)2 = k h=1 p 2α h h (p h 1)+2 p h +1 Definition Q = k p h +1 h=1 p h 1 Lemma φ(c n ) > n2 Q

44 An inequality If G is a counter example to main theorem: average of φ(o(g)) over G:

45 An inequality If G is a counter example to main theorem: average of φ(o(g)) over G: φ(g) n φ(cn) n > n Q

46 An inequality If G is a counter example to main theorem: average of φ(o(g)) over G: φ(g) n φ(cn) n > n Q g G with n < Qφ(o(g))

47 An inequality If G is a counter example to main theorem: average of φ(o(g)) over G: φ(g) n φ(cn) n > n Q g G with n < Qφ(o(g)) Key Theorem (technical proof) p : largest prime divisor of n If g G\{id} st n < Qφ(o(g)), (as occurs if counterexample) Then normal (unique) Sylow p-subgroup P of G. P g, so P cyclic.

48 Structure Theorem (Schur-Zassenhaus) If K G with ( K, G : K ) = 1, then

49 Structure Theorem (Schur-Zassenhaus) If K G with ( K, G : K ) = 1, then G = K ϕ H (semidirect product) for some H G and some homomorphism ϕ : H Aut(K).

50 Structure Theorem (Schur-Zassenhaus) If K G with ( K, G : K ) = 1, then G = K ϕ H (semidirect product) for some H G and some homomorphism ϕ : H Aut(K). Corollary If g G\{id} st n < Qφ(o(g)): G = P ϕ H (semidirect product) P cyclic sylow p-group H subgroup with P, H coprime.

51 Semidirect products Lemma K : finite abelian group H : finite group, ( K, H ) = 1

52 Semidirect products Lemma K : finite abelian group H : finite group, ( K, H ) = 1 k K, h H o K ϕh(kh) o K H (kh) (order in semi direct product divides order in direct product)

53 Semidirect products Lemma K : finite abelian group H : finite group, ( K, H ) = 1 k K, h H o K ϕh(kh) o K H (kh) (order in semi direct product divides order in direct product) Corollary φ(k ϕ H) φ(k H).

54 Semidirect products Lemma K : finite abelian group H : finite group, ( K, H ) = 1 k K, h H o K ϕh(kh) o K H (kh) (order in semi direct product divides order in direct product) Corollary Lemma φ(k ϕ H) φ(k H). φ(k H) φ(k)φ(h)

55 Semidirect products Lemma K : finite abelian group H : finite group, ( K, H ) = 1 k K, h H o K ϕh(kh) o K H (kh) (order in semi direct product divides order in direct product) Corollary Lemma φ(k ϕ H) φ(k H). φ(k H) φ(k)φ(h) equality when ( K, H ) = 1.

56 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct.

57 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g.

58 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd.

59 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd. Suppose h generates H but not J.

60 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd. Suppose h generates H but not J. So m = o J (h) = n/2 is odd.

61 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd. Suppose h generates H but not J. So m = o J (h) = n/2 is odd. Now L = h J, L = m odd, J : L = 2.

62 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd. Suppose h generates H but not J. So m = o J (h) = n/2 is odd. Now L = h J, L = m odd, J : L = 2. Let K be a Sylow 2-subgroup of G, so K = 2.

63 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd. Suppose h generates H but not J. So m = o J (h) = n/2 is odd. Now L = h J, L = m odd, J : L = 2. Let K be a Sylow 2-subgroup of G, so K = 2. Now J = LK and L K = {id}, so J = L ψ K = C m ψ C 2.

64 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd. Suppose h generates H but not J. So m = o J (h) = n/2 is odd. Now L = h J, L = m odd, J : L = 2. Let K be a Sylow 2-subgroup of G, so K = 2. Now J = LK and L K = {id}, so J = L ψ K = C m ψ C 2. Since J not cylic, it is dihedral group D 2m.

65 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd. Suppose h generates H but not J. So m = o J (h) = n/2 is odd. Now L = h J, L = m odd, J : L = 2. Let K be a Sylow 2-subgroup of G, so K = 2. Now J = LK and L K = {id}, so J = L ψ K = C m ψ C 2. Since J not cylic, it is dihedral group D 2m. C m 0 = C m 0 same cogenerators of subgroups.

66 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd. Suppose h generates H but not J. So m = o J (h) = n/2 is odd. Now L = h J, L = m odd, J : L = 2. Let K be a Sylow 2-subgroup of G, so K = 2. Now J = LK and L K = {id}, so J = L ψ K = C m ψ C 2. Since J not cylic, it is dihedral group D 2m. C m 0 = C m 0 same cogenerators of subgroups. C m 0, C m 0 can t cogenerates w/ C m 1, C m 1, resp.

67 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd. Suppose h generates H but not J. So m = o J (h) = n/2 is odd. Now L = h J, L = m odd, J : L = 2. Let K be a Sylow 2-subgroup of G, so K = 2. Now J = LK and L K = {id}, so J = L ψ K = C m ψ C 2. Since J not cylic, it is dihedral group D 2m. C m 0 = C m 0 same cogenerators of subgroups. C m 0, C m 0 can t cogenerates w/ C m 1, C m 1, resp. C m 1 flips, no cogenerators, C m 1 has all generators of C 2m

68 Special case Lemma. a, b : coprime positive integers J := φ(c a ϕ C b ) φ(c a C b ) =: H equality iff the semi-direct product is direct. Suppose equality holds. Then φ(o J (g)) = φ(o H (g)) g. Now o J (g) = o H (g) or o J (g) = 2o H (g) with o H (g) odd. Suppose h generates H but not J. So m = o J (h) = n/2 is odd. Now L = h J, L = m odd, J : L = 2. Let K be a Sylow 2-subgroup of G, so K = 2. Now J = LK and L K = {id}, so J = L ψ K = C m ψ C 2. Since J not cylic, it is dihedral group D 2m. C m 0 = C m 0 same cogenerators of subgroups. C m 0, C m 0 can t cogenerates w/ C m 1, C m 1, resp. C m 1 flips, no cogenerators, C m 1 has all generators of C 2m equality fails, contradiction unless s.d.p. is direct.

69 Outline of proof of main result Strategy Induct on number of prime factors.

70 Outline of proof of main result Strategy Induct on number of prime factors. Launch with prime powers counterexample would be cyclic.

71 Outline of proof of main result Strategy Induct on number of prime factors. Launch with prime powers counterexample would be cyclic. If G counterexample: G = P ϕ H

72 Outline of proof of main result Strategy Induct on number of prime factors. Launch with prime powers counterexample would be cyclic. If G counterexample: G = P ϕ H P cyclic sylow p-group, ( P, H ) = 1, H fewer prime divors than G

73 Outline of proof of main result Strategy Induct on number of prime factors. Launch with prime powers counterexample would be cyclic. If G counterexample: G = P ϕ H P cyclic sylow p-group, ( P, H ) = 1, H fewer prime divors than G φ(g) φ(p H) = φ(p)φ(h)

74 Outline of proof of main result Strategy Induct on number of prime factors. Launch with prime powers counterexample would be cyclic. If G counterexample: G = P ϕ H P cyclic sylow p-group, ( P, H ) = 1, H fewer prime divors than G φ(g) φ(p H) = φ(p)φ(h) φ(h) φ(c H ), equal iff H cyclic

75 Outline of proof of main result Strategy Induct on number of prime factors. Launch with prime powers counterexample would be cyclic. If G counterexample: G = P ϕ H P cyclic sylow p-group, ( P, H ) = 1, H fewer prime divors than G φ(g) φ(p H) = φ(p)φ(h) φ(h) φ(c H ), equal iff H cyclic H must be cyclic

76 Outline of proof of main result Strategy Induct on number of prime factors. Launch with prime powers counterexample would be cyclic. If G counterexample: G = P ϕ H P cyclic sylow p-group, ( P, H ) = 1, H fewer prime divors than G φ(g) φ(p H) = φ(p)φ(h) φ(h) φ(c H ), equal iff H cyclic H must be cyclic G = C P ϕ C H & C n = C P C H

77 Outline of proof of main result Strategy Induct on number of prime factors. Launch with prime powers counterexample would be cyclic. If G counterexample: G = P ϕ H P cyclic sylow p-group, ( P, H ) = 1, H fewer prime divors than G φ(g) φ(p H) = φ(p)φ(h) φ(h) φ(c H ), equal iff H cyclic H must be cyclic G = C P ϕ C H & C n = C P C H φ(c P ϕ φ(c H ) = φ(c P C H ) iff s.d.p. is direct

78 Outline of proof of main result Strategy Induct on number of prime factors. Launch with prime powers counterexample would be cyclic. If G counterexample: G = P ϕ H P cyclic sylow p-group, ( P, H ) = 1, H fewer prime divors than G φ(g) φ(p H) = φ(p)φ(h) φ(h) φ(c H ), equal iff H cyclic H must be cyclic G = C P ϕ C H & C n = C P C H φ(c P ϕ φ(c H ) = φ(c P C H ) iff s.d.p. is direct contradiction; no counterexamples.

79 A group sum inequality and its application to power graphs To appear Bulletin of the Australian Mathematical Society

80 A group sum inequality and its application to power graphs To appear Bulletin of the Australian Mathematical Society ArXiv arxiv:

81 A group sum inequality and its application to power graphs To appear Bulletin of the Australian Mathematical Society ArXiv arxiv: Thank you