#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

1. Find P2 * P1. Are they the same?

P2 * P1 = 32154. Thus P1 * P2 and P2 * P1 are not the same.

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

If the order of p is k, then that means pi^k is the identity permutation.
The function GG() returns all of the permutations of some permutation group,
and thus there are only k members in GG, and thus |GG({pi})| = k.

3. Prove that if H is a subgroup of a group G, then |G|/|H| is an integer.

The left cosets of H in G are the equivalence classes of a certain equivalence relation on G.
Therefore, the left cosets form a partition of G. Each left coset aH has the same cardinality
as H because x--> ax defines a bijection H--> aH. Therefore if H is a subgroup of a group G,
then |G|/|H| is an integer.

4. Whose theorem is it?

Lagrange

5. Generate a random permutation of length 9 (use randperm(9)) and by hand find its cycle representation. Check it with PtoC(pi).

randperm(9) --> [8,5,6,4,3,9,7,2,1]

(9,1,8,2,5,3,6)(4)(7)

PtoC(%) --> [[4],[7],[9,1,8,2,5,3,6]]

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 with length 7)

#Not exactly sure what this question is asking
evalb(permute(FunT(PtoC(randperm(7)))) = permute(7)) --> true