src/HOL/Algebra/Exponent.thy
 author wenzelm Tue Oct 10 19:23:03 2017 +0200 (2017-10-10) changeset 66831 29ea2b900a05 parent 66453 cc19f7ca2ed6 permissions -rw-r--r--
tuned: each session has at most one defining entry;
1 (*  Title:      HOL/Algebra/Exponent.thy
2     Author:     Florian Kammueller
3     Author:     L C Paulson
5 exponent p s   yields the greatest power of p that divides s.
6 *)
8 theory Exponent
9 imports Main "HOL-Computational_Algebra.Primes"
10 begin
12 section \<open>Sylow's Theorem\<close>
14 text \<open>The Combinatorial Argument Underlying the First Sylow Theorem\<close>
16 text\<open>needed in this form to prove Sylow's theorem\<close>
17 corollary (in algebraic_semidom) div_combine:
18   "\<lbrakk>prime_elem p; \<not> p ^ Suc r dvd n; p ^ (a + r) dvd n * k\<rbrakk> \<Longrightarrow> p ^ a dvd k"
19   by (metis add_Suc_right mult.commute prime_elem_power_dvd_cases)
21 lemma exponent_p_a_m_k_equation:
22   fixes p :: nat
23   assumes "0 < m" "0 < k" "p \<noteq> 0" "k < p^a"
24     shows "multiplicity p (p^a * m - k) = multiplicity p (p^a - k)"
25 proof (rule multiplicity_cong [OF iffI])
26   fix r
27   assume *: "p ^ r dvd p ^ a * m - k"
28   show "p ^ r dvd p ^ a - k"
29   proof -
30     have "k \<le> p ^ a * m" using assms
31       by (meson nat_dvd_not_less dvd_triv_left leI mult_pos_pos order.strict_trans)
32     then have "r \<le> a"
33       by (meson "*" \<open>0 < k\<close> \<open>k < p^a\<close> dvd_diffD1 dvd_triv_left leI less_imp_le_nat nat_dvd_not_less power_le_dvd)
34     then have "p^r dvd p^a * m" by (simp add: le_imp_power_dvd)
35     thus ?thesis
36       by (meson \<open>k \<le> p ^ a * m\<close> \<open>r \<le> a\<close> * dvd_diffD1 dvd_diff_nat le_imp_power_dvd)
37   qed
38 next
39   fix r
40   assume *: "p ^ r dvd p ^ a - k"
41   with assms have "r \<le> a"
42     by (metis diff_diff_cancel less_imp_le_nat nat_dvd_not_less nat_le_linear power_le_dvd zero_less_diff)
43   show "p ^ r dvd p ^ a * m - k"
44   proof -
45     have "p^r dvd p^a*m"
46       by (simp add: \<open>r \<le> a\<close> le_imp_power_dvd)
47     then show ?thesis
48       by (meson assms * dvd_diffD1 dvd_diff_nat le_imp_power_dvd less_imp_le_nat \<open>r \<le> a\<close>)
49   qed
50 qed
52 lemma p_not_div_choose_lemma:
53   fixes p :: nat
54   assumes eeq: "\<And>i. Suc i < K \<Longrightarrow> multiplicity p (Suc i) = multiplicity p (Suc (j + i))"
55       and "k < K" and p: "prime p"
56     shows "multiplicity p (j + k choose k) = 0"
57   using \<open>k < K\<close>
58 proof (induction k)
59   case 0 then show ?case by simp
60 next
61   case (Suc k)
62   then have *: "(Suc (j+k) choose Suc k) > 0" by simp
63   then have "multiplicity p ((Suc (j+k) choose Suc k) * Suc k) = multiplicity p (Suc k)"
64     by (subst Suc_times_binomial_eq [symmetric], subst prime_elem_multiplicity_mult_distrib)
65        (insert p Suc.prems, simp_all add: eeq [symmetric] Suc.IH)
66   with p * show ?case
67     by (subst (asm) prime_elem_multiplicity_mult_distrib) simp_all
68 qed
70 text\<open>The lemma above, with two changes of variables\<close>
71 lemma p_not_div_choose:
72   assumes "k < K" and "k \<le> n"
73       and eeq: "\<And>j. \<lbrakk>0<j; j<K\<rbrakk> \<Longrightarrow> multiplicity p (n - k + (K - j)) = multiplicity p (K - j)" "prime p"
74     shows "multiplicity p (n choose k) = 0"
75 apply (rule p_not_div_choose_lemma [of K p "n-k" k, simplified assms nat_minus_add_max max_absorb1])
76 apply (metis add_Suc_right eeq diff_diff_cancel order_less_imp_le zero_less_Suc zero_less_diff)
77 apply (rule TrueI)+
78 done
80 proposition const_p_fac:
81   assumes "m>0" and prime: "prime p"
82   shows "multiplicity p (p^a * m choose p^a) = multiplicity p m"
83 proof-
84   from assms have p: "0 < p ^ a" "0 < p^a * m" "p^a \<le> p^a * m"
85     by (auto simp: prime_gt_0_nat)
86   have *: "multiplicity p ((p^a * m - 1) choose (p^a - 1)) = 0"
87     apply (rule p_not_div_choose [where K = "p^a"])
88     using p exponent_p_a_m_k_equation by (auto simp: diff_le_mono prime)
89   have "multiplicity p ((p ^ a * m choose p ^ a) * p ^ a) = a + multiplicity p m"
90   proof -
91     have "(p ^ a * m choose p ^ a) * p ^ a = p ^ a * m * (p ^ a * m - 1 choose (p ^ a - 1))"
92       (is "_ = ?rhs") using prime
93       by (subst times_binomial_minus1_eq [symmetric]) (auto simp: prime_gt_0_nat)
94     also from p have "p ^ a - Suc 0 \<le> p ^ a * m - Suc 0" by linarith
95     with prime * p have "multiplicity p ?rhs = multiplicity p (p ^ a * m)"
96       by (subst prime_elem_multiplicity_mult_distrib) auto
97     also have "\<dots> = a + multiplicity p m"
98       using prime p by (subst prime_elem_multiplicity_mult_distrib) simp_all
99     finally show ?thesis .
100   qed
101   then show ?thesis
102     using prime p by (subst (asm) prime_elem_multiplicity_mult_distrib) simp_all
103 qed
105 end