#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



#ATTENDANCE Q. #1 for LECTURE 10
#Find P2*P1, are they the same?

#ANSWER to Q. #1:	

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 was 14523, and P2*P1 is 32154, and so they are not the same




#ATTENDANCE Q. #2 for LECTURE 10
#In your own words, explain why if the order of pi is k then |GG({pi})| = k

#ANSWER to Q. #2:	This is likely because GG achieves all the unique subgroups, which can involve
#			going through all of the different subgroup permutations such that those
#			subgroups do satisfy the conditions of being a group.  Then, it is possible
#			to go through the cycle of all those unique subgroups, but at some point
#			you'll end up at the identity permutation again, and after that there are
#			no more unique permutations (since you're starting a functional cycle again),
#			and so it gives the same number as the Order of pi, which is the smallest
#			number m for which (pi)^m is the identity permutation.




#ATTENDANCE Q. #3 for LECTURE 10
#Prove that if H is a subgroup of a group G, then |G|/|H| is always an integer (can look up)

#ANSWER to Q. #3:	I found a proof on Wikipedia, and overall it establishes an equivalence
#			relation with particular elements in H and G such that (x and y in G, h
#			in H), x=yh.  Then these "left cosets" form a partition of G, and then
#			it is true that each left coset aH has the same cardinality as H because
#			of the bijection H --> aH from x --> ax, and then they conclude that
#			|G| = [G:H]*|H| where the index [G:H] is the number of left cosets.




#ATTENDANCE Q. #4 for LECTURE 10
#Whose theorem is it?

#ANSWER to Q. #4:	This is Lagrange's theorem.




#ATTENDANCE Q. #5 for LECTURE 10
#Find cycle representation of a randperm(9), check with PtoC(pi)

#ANSWER to Q. #5:	 1  2  3  4  5  6  7  8  9
#			[3, 8, 4, 9, 1, 5, 6, 7, 2]
#			
#			1->3->4->9->2->8->7->6->5->1  CYCLES are (134928765)
#			This is the same as PtoC(pi).





#ATTENDANCE Q. #6 for LECTURE 10
#Check that the range of FunT over all permutations of length 7 is the same thing as 
#the set of permutations of length 7

#ANSWER to Q. #6:	Yes, they are the same. This can be found by writing a quick proc
#			that adds elements FunT(permute(7)) into a set, and then finding
#			the set subtraction of that set minus the set of permutations of
#			length 7, to get {} as the answer, meaning they have the same items.