modulo multiplication is group or not with proof
Answers
Step-by-step explanation:
A group is a set, together with a binary operation ⊙, such that the following conditions hold:
For all a, b ∈ S it holds that a ⊙ b ∈ S
Now, i have to show that the Set = {1, 2 ... 16, 17} together with multiplication modulo 18 is not a group.
Answer:
A modulo multiplication group is a finite group M_m of residue classes prime to m under multiplication mod m. M_m is Abelian of group order phi(m), where phi(m) is the totient function.
ModuloMultiplicationGroups
A modulo multiplication group can be visualized by constructing its cycle graph. Cycle graphs are illustrated above for some low-order modulo multiplication groups. Such graphs are constructed by drawing labeled nodes, one for each element A of the residue class, and connecting cycles obtained by iterating A^n. Each edge of such a graph is bidirected, but they are commonly drawn using undirected edges with double edges used to indicate cycles of length two (Shanks 1993, pp. 85 and 87-92).
The following table gives the modulo multiplication groups of small orders, together with their isomorphisms with respect to cyclic groups C_n.
M_m group phi(m) elements
M_2 <e> 2 1
M_3 C_2 2 1, 2
M_4 C_2 2 1, 3
M_5 C_4 4 1, 2, 3, 4
M_6 C_2 2 1, 5
M_7 C_6 6 1, 2, 3, 4, 5, 6
M_8 C_2×C_2 4 1, 3, 5, 7
M_9 C_6 6 1, 2, 4, 5, 7, 8
M_(10) C_4 4 1, 3, 7, 9
M_(11) C_(10) 10 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
M_(12) C_2×C_2 4 1, 5, 7, 11
M_(13) C_(12) 12 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
M_(14) C_6 6 1, 3, 5, 9, 11, 13
M_(15) C_2×C_4 8 1, 2, 4, 7, 8, 11, 13, 14
M_(16) C_2×C_4 8 1, 3, 5, 7, 9, 11, 13, 15
M_(17) C_(16) 16 1, 2, 3, ..., 16
M_(18) C_6 6 1, 5, 7, 11, 13, 17
M_(19) C_(18) 18 1, 2, 3, ..., 18
M_(20) C_2×C_4 8 1, 3, 7, 9, 11, 13, 17, 19
M_(21) C_2×C_6 12 1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20
M_(22) C_(10) 10 1, 3, 5, 7, 9, 13, 15, 17, 19, 21
M_(23) C_(22) 22 1, 2, 3, ..., 22
M_(24) C_2×C_2×C_2 8 1, 5, 7, 11, 13, 17, 19, 23
M_m is a cyclic group (which occurs exactly when m has a primitive root) iff m is of one of the forms m=2, 4, p^n, or 2p^n, where p is an odd prime and n>=1 (Shanks 1993, p. 92). The first few of these are m=3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 17, 18, 19, ... (OEIS A033948; Shanks 1993, p. 84).
The only ordered m for which the elements of M_m are all self-conjugate are the divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24 (OEIS A018253; Eggar 2000). These correspond to the groups <e>, C_2, C_2×C_2, and C_2×C_2×C_2. This also means that no modulo multiplication group is isomorphic to a direct product of more than three copies of C_2.
Isomorphic modulo multiplication groups can be determined using a particular type of factorization of the totient function phi(m) using the property that
phi(p^alpha)=p^(alpha-1)(p-1)
(1)
as described by Shanks (1993, pp. 92-93). To perform this factorization, begin by analogy with computation of the totient function by factoring m in the standard form
m=p_1^(a_1)p_2^(a_2)...p_n^(a_n).
(2)
Now for each power of an odd prime, write
phi(p_i^(a_i))=(p_i-1)p_i^(a_i-1),
(3)
and factor the leading term
p_i-1=q_1^(b_1)q_2^(b_2)...q_s^(b_s)
(4)
as
<q_1^(b_1)><q_2^(b_2)>...<q_s^(b_s)><p_i^(a_i-1)>,
(5)
where <q^b> denotes the explicit expansion of q^b (i.e., 5^2=25), and the last term is omitted if a_i=1 (since in that case, <p_i^(a_i-1)>=1).
If m contains a power of 2 so that p_1=2, then write
phi(2^(a_1))={nothing for a_1<2; <2> for a_1=2; <2><2^(a_1-2)> for a_1>2.
(6)
Now combine terms from the odd and even primes, write them as a product and combine any unambiguous products of terms. The resulting expression is denoted phi_m and the group M_m is isomorphic to a direct product of cyclic groups of orders given by phi_m.
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