#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

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Attendance Problem 1:
Find P2*P1. Are they the same?

P1 = 21534, P2 = 41235
P2*P1 = 32154 which does not equal 14523 so they are not the same.

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

GG{pi} is the subgroup generated by the set {pi}. This subgroup is generated by starting with pi and multiplying by pi over and over until you reach the original permutation you started with. Since we know the order of pi is k, that means pi can be multiplied by itself k times before getting back the original permutation. So, it makes sense that |GG{pi}| = k because there are k distinct permutations obtained before you get back to the original permutation and repeat.


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

(I haven't taken any classes involving group theory yet so I'm not sure about some of the terminology.)

Let H be a subgroup of the group G and let g be an element of G. The set gH is a left coset of H in G. We know by the cancellation property that |gH| = |H| and so every coset of H has the same order. We can also show that for any two left cosets, those cosets are either the same or are disjoint. 

So, we see that the cosets partition G into mutually disjoint subsets which each have |H| elements meaning that |G| / |H| has to be an integer since we can only have an integer number of parts in a partition.


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Attendance Problem 4:
Whose theorem is it?

It was Joseph-Louis Lagrange's Theorem.

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Attendance Problem 5:
Generate a random permutation of length 9 (use randperm(9))
Find its cycle representation by hand, then check it with PtoC(pi).

[8,5,6,4,3,9,7,2,1] is the random permutation.

1 -> 8 -> 2 -> 5 -> 3 -> 6 -> 9 -> 1 
Largest number is 9 so (9182536)
4 -> 4
(4)
7 -> 7
(7)

The cycle representation is (4)(7)(9182536).

PtoC([8,5,6,4,3,9,7,2,1]) = [[4], [7], [9, 1, 8, 2, 5, 3, 6]] which is what we found by hand.


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Attendance Problem 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).

{seq(FunT(permute(7)[i]),i=1..nops(permute(7)))} minus {op(permute(7))} 

Returns the empty set which means that the range of FunT over all permutations of length 7 is the same thing as the set of permutation of length 7.