#ATTENDANCE QUIZ FOR LECTURE 10 of Dr. Z.'s Math454(02) Rutgers University

# Please Edit this .txt page with Answers

#Email  ShaloshBEkhad@gmail.com 
#Subject: p10
#with an attachment called
#p10FirstLast.txt
#(e.g. p10DoronZeilberger.txt)

#Right after finishing watching the lecture but no later than Oct. 9, 2020, 8:00pm


THE NUMBER OF ATTENDANCE QUESTIONS WERE: 6

PLEASE LIST ALL THE QUESTIONS FOLLOWED, AFTER EACH BY THE ANSWER

#Question 1: Find P2 * P1. Are P1 * P2 and P2 * P1 the same? 

#Answer 1: 
# P1 = 21534, P2 = 41235. 
# 1 2 3 4 5	1 2 3 4 5     1 2 3 4 5
#[         ] * [         ] = [         ]
# 4 1 2 3 5     2 1 5 3 4     3 2 1 5 4

# P1 * P2 = 14523. I got P2 * P1 to be 32154. P1 * P2 does not equal P2 * P1. 

# Question 2: Explain in your own words why if the order of pi is k then |GG({pi})| = k

# Answer 2: 
# When the order of pi is k it means the group is
# e, pi, pi^2, ...., pi^k, so it makes sense that |GG({pi})|, which is the number of elements in the group, is k. 

# Question 3: Prove that if H is a subgroup of a group G then: |G|/|H| is an integer

# Answer 3:  
# A left/right coset is the set of when you multiply all elements of H on the left/right by one element of G.
# G consists of disjoint cosets with |H| elements as you are just multiplying the elements of H. 
# Since the union of the left cosets of H in G is G and they are disjoint, they partition G. 
# These cosets perfectly partition G, so |H| must indeed be a divisor of |G|. 

# Question 4: Whose theorem is it? 

# Answer 4: It is Lagrange's Theorem. 

# Question 5: Generate a random permutation of length 9 (use randperm(9)) and by hand find its cycle representation.

# Answer 5: The permutation I got is , 

# 1 2 3 4 5 6 7 8 9
# 2 3 6 1 4 5 8 9 7

# 1 --> 2 --> 3 --> 6 --> 5 --> 4 --> 1 
# 7 --> 8 --> 9 --> 7

# The cycles are (123654) and (789). 

# Question 6: Check that the range of FunT over all permutations of length 7 (you can use permute(7)) is the same thing (the set of permutations of length 7).

# Answer 6: 
check := proc(p): 
local i, ans:
ans := {}:
for i to nops(p) do
  ans := ans union {FunT(p[i])}:
od:
return ans:
end proc:

nops(permute(7));
#                              5040

nops(check(permute(7)));
#                              5040

They are equal.