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

# Please Edit this .txt page with Answers#Email: hra25@scarletmail.rutgers.edu
#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:

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

#Q1 Find P2*P1, are they the same?

# P2*P1 = [1, 2, 3, 4, 5; 3, 2, 5, 1, 4], which is not the same as P1*P2

#Q2 Explain in your own words why if the order of pi is k, then |GG({pi})|=k

# GG() permutes and iterates through the set (given as a param) in order to generate the desired result; it follows from this that it is based on the number of orders. Since there are k terms in the set, we have |GG({pi})| = k.

#Q3 Prove that if H is a subgroup of a group G, then |G|/|H| is an interger.

#A: Since we can take answers from online, one proof for lagranges Theorem is "For any element x of G, Hx = {h • x | h is in H} defines a right coset of H. By the cancellation law each h in H will give a different product when multiplied on the left onto x. Thus each element of H will create a corresponding unique element of Hx. Thus Hx will have the same number of elements as H. " (taken from http://dogschool.tripod.com/lagrange.html).

# Proof: Let x in G be arbitrary. By the Cancellation Law, hx is unique for each element h in H. Therefore, each unique element of H will generate a unique element of Hx (the right coset of H under x). Therefore, the order of Hx is equal to that of H, as required. QED

#Q4 Whose theorem is it?

#A: This is Lagrange's Theorem in group theory

#Q5 Generate a random permuation of length 9 (use randperm(9)) and by hand find its cycle representation and check it with PtoC(pi) (1:05:00)

                    [1,2,3,4,5,6,7,8,9]
# S:=randperm(9) -> [8,5,6,4,3,9,7,2,1]
# The cyclic representation of S is {[8, 2, 5, 3, 6, 9, 1], [4], [7]}, which is verified by PtoC(pi).

#Q6 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) (1:24:00)

#A: FunT(permute(7)) does not halt in Maple -- therefore, it is not possible to verify this.