# Disjoint Cycles

• Jul 19th 2012, 08:26 PM
Magical
Disjoint Cycles
P=
(1234567
3125746)

q=
(1234567
5641723)

r=
(1234567
7631425)

s=
(1234567
5713264)

Find pq(^-1)sr(^2) write the answer as a product of disjoint cylcles. Find its order Parity and Inverse.

I understand parts of the question but for the majority I am unable to do it.
• Jul 20th 2012, 06:38 PM
beebe
Re: Disjoint Cycles
Which parts, specifically, are you unable to do?
• Jul 20th 2012, 08:47 PM
Magical
Re: Disjoint Cycles
(132)(4576)*(15734)(26)
(1
q sends 1 to 5 and p sends 5 to 7
(17- because p.q(1) = p(5) = 7
so the composite permutation begins (17...)
then see what p.q(7) is and so on
q sends 7 to 3 and p sends 3 to 2
(172
q sends 2 to 6 and p sends 6 to 4
(1724
q sends 4 to 1 and p sends 1 to 3
(17243
q sends 3 to 4 and p sends 4 to 5
(172435
q sends 5 to 7 and p sends 7 to 6
(1724356
just to check, q sends 6 to 2 and p sends 2 to 1
(1724356)-7 digits so completed as pq

To find sr we do the same
Sr
1
R takes 1 to 7 and s takes 7 to 4
14
R takes 4 to 1 and s takes 1 to 5
145
R takes 5 to 4 and s takes 4 to 3
1453
R takes 3 to 3 and s takes 3 to 1
14531

this is as far as ive got
• Jul 21st 2012, 12:35 AM
Deveno
Re: Disjoint Cycles
ok, well, i do not know if for you:

pq means first do q, and then do p, or vice-versa (different authors use different conventions). i will assume pq(x) = p(q(x)) (do q first).

so first we have to calculate pq-1sr2. this is tedious.

first we do r twice:

1→7→5
2→6→2
3→3→3
4→1→7
5→4→1
6→2→6
7→5→4

then we apply s:

1→5→2
2→2→7
3→3→1
4→7→4
5→1→5
6→6→6
7→4→3

then we apply q-1, which is:

1→4
2→6
3→7
4→3
5→1
6→2
7→5, to get:

1→2→6
2→7→5
3→1→4
4→4→3
5→5→1
6→6→2
7→3→7

finally, we apply p to get:

1→6→4
2→5→7
3→4→5
4→3→2
5→1→3
6→2→1
7→7→6, so that pq-1sr2 = (1 4 2 7 6)(3 5)

this is the product of a disjoint 5-cycle and 2-cycle, so has order lcm(2,5) = 10.

alternatively, we have pq-1sr2 = (1 3 2)(4 5 7 6)(1 4 3 7 5)(2 6)(1 5 2 7 4 3)(1 7 5 4)(2 6)(1 7 5 4)(2 6).

thus pq-1sr2(1) = p(q-1(s(r(r(1)))))

= p(q-1(s(r(7)))) = p(q-1(s(5))) = p(q-1(2)) = p(6) = 4 so we start:

(1 4......

then we calculate pq-1sr2(4) = p(q-1(s(r(r(4)))))

= p(q-1(s(r(1)))) = p(q-1(s(7))) = p(q-1(4)) = p(3) = 2, so we continue:

(1 4 2.....

next, we want to know pq-1sr2(2) = p(q-1(s(r(r(2)))))

= p(q-1(s(r(6)))) = p(q-1(s(2))) = p(q-1(7)) = p(5) = 7, so we are up to:

(1 4 2 7....

AND.....pq-1sr2(7) = p(q-1(s(r(r(7)))))

= p(q-1(s(r(5)))) = p(q-1(s(4))) = p(q-1(3)) = p(7) = 6, so we have:

(1 4 2 7 6.....

and quickly, now! pq-1sr2(6) = p(q-1(s(r(r(6)))))

= p(q-1(s(r(2)))) = p(q-1(s(6))) = p(q-1(6)) = p(2) = 1, so our first cycle closes:

(1 4 2 7 6)(....

the next number not in the first cycle is 3, so we calculate:

pq-1sr2(3) = p(q-1(s(r(r(3)))))

= p(q-1(s(r(3)))) = p(q-1(s(3))) = p(q-1(1)) = p(4) = 5, and by bijectivity,

we must have pq-1sr2(5) = 3, so we get (1 4 2 7 6)(3 5), as we did above.

5-cycles are even, and (3 5) is a transposition, therefore pq-1sr2 is odd.