Dr. Marques Sophie
Office 519
Algebra 1
Fall Semester 2014
marques@cims.nyu.edu
Problem Set 9
Exercise 1 :
In Sn with n “ 10 :
(a) Find a permutation a such that axa´1 “ y if x “ p1, 2qp3, 4q and
y “ p5, 6qp3, 1q
(b) Explain why there is no element a such that ap1, 2, 3qa´1 “
p1, 3qp5, 7, 8q.
(c) Explain why there is no element a such that ap1, 2qa´1 “
p3, 4qp1, 5q.
Solution :
Both (b) and 3. follow because conjugation cannot change the cycle type of a permutation. For (a), we need a permutation τ such that τxτ´1 “ τp1, 2qτ´1 ¨ τp3, 4qτ´1
equals to p5, 6qp3, 1q. This happens if τp1, 2qτ´1 “ pτp1q, τp2qq “ p5, 6q and τp3, 4qτ´1 “
pτp3q, τp4qq “ p3, 1q, so we want τp1q “ 5 and τp2q “ 6, τp3q “ 3, τp4q “ 1. We can
send the remaining points r1, 10s „ t1, 2, 3, 4u to the six points in r1, 10s „ t1, 3, 5, 6u
any way we wish (6! choices) without affecting the desired identity τxτ´1 “ y.
Exercise 2 :
Prove that the cycles σ “ p1, 2q and τ “ p1, 2, . . . , nq generate Sn .
Hint : It would suffice to show that every 2-cycle can be written as a word in the letters
σ, τ and their inverses. Start by computing conjugates τστ´1 , τ2 στ´2 , . . ..
Solution :
Given σ “ p1, 2q, τ “ p1, 2, 3, . . . , nq, show ă σ, τ ą includes all 2´cycles since every
x P Sn is some product of 2-cycles. Let’s see what we get conjugating σ with τ ; note that
τ “ p1, n, n ´ 1, . . . , 2q but we don’t really need this. From our discussion of conjugation
in Sn :
τp1, 2qτ´1 “ pτp1q, τp2qq “ p2, 3q
τp2, 3qτ´1 “ pτp2q, τp3qq “ p3, 4q
and
τpn ´ 1, nqτ´1 “ pτpn ´ 1q, τpnqq “ pn, 1q “ p1, nq
Thus, H contains all 2-cycles of the form p1, 2q, p2, 3q, . . . , pn, 1q is all ”flips” of adjacent
entries in t1, 2, . . . , nu, including the end-flips pn, 1q.
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Recalling that xpi1 , . . . , ir qx´1 “ pxpi1 q, . . . , xpir qq, we might next try to get x so xp1, 2qx´1 “
p1, 3q ; that’s easy, just take x “ p2, 3q (which is in H). Thus, p1, 3q P H. Then
xp1, 3qx´1 “ p1, 4q if we take x “ p3, 4q, which is in H. Finally, we see that p1, n´1q P H
and then p1, nqp1, n ´ 1qp1, nq “ p1, nq. Hence all 2-cycles p1, kq, 2 ď k ď n are in H.
Once we get this far, try x “ p1, jq acting upon p1, iq by conjugation. For i ‰ j, we get :
xp1, iqx´1 “ pxp1q, xpiqq “ p j, iq “ pi, jq P H
with x “ p1, jq P H. Thus H contains all 2-cycles, and hence H “ G.
Exercise 3 :
Express the following permutations in S10 as products of commuting disjoint cycles, and
determine the parity of each operator.
(a) (1,2,3)(1,2) (b) (1,2,3,4,5)(1,2,3)(4,5) (c) (1,2)(1,3)(1,4)(2,5)
Solution :
paq σ “ p1, 2, 3qp1, 2q. Here sgnp1, 2, 3q “ 1, sgnp1, 2q “ ´1 so σ is odd.
pbq σ “ p1, 2, 3, 4, 5qp1, 2, 3qp4, 5q has sgn “ p`1qp`1qp´1q “ ´1 so it is odd.
pcq σ “ p1, 2qp1, 3qp1, 4qp1, 5q has sgn “ p´1q4 “ `1 and is even.
Exercise 4 :
For n ě 3 prove that the center
ZpSn q “ tσ P Sn : τσ “ στ for all τ P Sn u
is trivial. Where did you use the hypothesis that n ě 3 in your discussion ?
Hint : If σ ‰ e there are indices i ‰ j such that σpiq “ j. If σ P ZpSn q then
σpi, jqσ´1 “ pσpiq, σpjqq would equal pi, jq for all i ‰ j.
Note : Sn – Z2 (abelian) when n “ 2.
Solution :
We argue by contradiction. If there is σ P ZpSn q such that σ ‰ e, there is elements
i ‰ j in r1, ns such that σpiq “ j. By relabeling elements of r1, ns, we assume i “ 1,
j “ 2, σp1q “ 2. Consider any k ‰ 1, 2 and the conjugate σp1, kqσ´1 “ pσp1q, σpkqq
which must equal p1, kq. Since σp1q “ 2 ‰ 1, we must have σp2q “ k. But this applies
to all k ‰ 1, 2. We have an immediate contradiction if there is more than one index
in r1, ns „ t1, 2u. The remaining case is r1, ns “ t1, 2, 3u in which case k mudt be 3.
σ /
σ /
Then 1
2
3 “ k . We can’t have σ : 3 Ñ 2 σ : 1 Ñ 2 because σ is 1 : 1.
Thus σp3q “ 1 and σ “ p1, 2, 3q in S3 . But this fails to commute with τ “ p1, 2q because
στσ´1 “ pσp1q, σp2qq “ p2, 3q is not equal to p1, 2q. In every case, we end up with a
contradiction.
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Exercise 5 :
The stabilizer of a permutation σ is the subgroup
StabSn pσq “ tx P Sn : xσx´1 “ σu
List all elements in the stabilizers of the following permutations
(a) σ “ p123q in S5
(b) σ “ p12qp345q in S5
What are the cardinalities of these stabilizers ?
Solution :
paq σ “ p1, 2, 3q in S5 . τ P S5 stabilizes σ if and only if the cycle ordered list pτp1q, τp2q, τp3qq
is the same 3-cycle as σ. That happens if and only if τp1q “ 1, τp1q “ 2, τp3q “ 3 ;
τ “ id on t1, 2, 3u with arbitrary action on the complementary set t4, 5u ; τp1q “ 2,
τp2q “ 3, τp3q “ 1 ; τ “ p1, 2, 3q on the set t1, 2, 3u and τp1q “ 3, τp2q “ 1, τp3q “ 2 :
τ “ p1, 3, 2q “ p1, 2, 3q´1 on t1, 2, 3u with arbitrary action on t4, 5u.
StabS5 pp2, 3qq “ te, p4, 5q, p1, 2, 3q, p1, 2, 3qp4, 5q, p1, 3, 2q, p1, 3, 2qp4, 5qu
This is the group generated by 2 commuting elements ă p1, 2, 3q, p4, 5q ą.
pbq σ “ p1, 2qp3, 4, 5q. Now, τ P StabS5 pσq must make τστ´1 “ τp1, 2qτ´1 ¨ τp3, 4, 5qτ´1 “
pτp1q, τp2qq ¨ pτp3q, τp4q, τp5qq equal tp p1, 2qp3, 4, 5q. By uniqueness of disjoint cycle decompositions that means pτp1q, τp2qq “ p1, 2q so τ “ Id on t1, 2u or τ “ p1, 2q on t1, 2u
while we also have pτp3q, τp4q, τp5qq “ p3, 4, 5q. On t3, 4, 5u, τ must be the Id on t3, 4, 5u,
or τ “ p3, 4, 5q on t3, 4, 5u or τ “ p3, 5, 4q on t3, 4, 5u. The full set of elements in the
stabilizer is
StabS5 pp1, 2qp3, 4, 5qq “ te, p3, 4, 5q, p3, 5, 4q “ p3, 4, 5q´1 ,
p1, 2q, p1, 2q, p1, 2qp3, 4, 5q, p1, 2qp3, 4, 5qu
This is precisely the group with 2-generators ă p1, 2q, p3, 4, 5q ą, that are disjoint commuting cycles.
Exercise 5 :
Prove that every element of Sn can be written as a product σ “ τ1 . . . τr of ”flips”
p1, 2q, p2, 3q, . . . , pn ´ 1, nq, pn, 1q
of adjacent elements in the cyclically ordered set r1, ns.
Hint : It suffices to show that any 2-cycle pi, jq is a product of ”flips”. (Why ?) Consider
conjugation operations on pi, jq.
Solution :
Since every σ P Sn can be written as a product of 2-cycles, it suffices to show any
2-cycles pi, jq, i ‰ j is a product of flips. Furthermore, by relabeling the elements in
r1, ns, it suffices to prove this for particular 2-cycle p1, kq, 1 ă k ď n.
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Conjugation σ Ñ τστ´1 of any cycle σ “ pi1 , . . . , ir q has the following result τστ´1 “
pτpi1 q, . . . , τpir qq “ another cycle of same length. Taking τ “ pk, k ´ 1q, we get
τp1, kqτ´1 “ pk, k ´ 1qp1, kqpk, k ´ 1q “ p1, k ´ 1q
(Note : If k “ 2 there is no need for a conjugation since p1, 2q is already a flip). We are
done if k ´ 1 “ 2, because we can rewrite this as
pk, k ´ 1qp1, kqpk, k ´ 1q “ p1, k ´ 1q
Then
p1, kq “ pk, k ´ 1qp1, k ´ 1qpk ´ 1q “ pk, k ´ 1qp1, 2qpk, k ´ 1q
a product of flips.
Otherwise, conjugate p1, k ´ 1q by τ “ pk ´ 2, k ´ 1q to get
p1, k´2q “ pk´2, k´1qpk´1, 1qpk´1, k´2q “ pk´2, k´1qpk´1, kqp1, kqpk´1, kqpk´2, k´1q
p1, kq “ pk ´ 2, k ´ 1qpk ´ 1, kqp1, k ´ 2qpk ´ 1, kqpk ´ 2, k ´ 1q
Continue by induction on k to get a final product of flips.
p1, kq “ p2, 3q . . . pk ´ 2, k ´ 1qpk ´ 1, kqpk ´ 2, k ´ 1q . . . p2, 3q
which is a product of flips.
Exercise 7 :
The matrix group G “ SLn pRq “ tA P Mn pRq : detpAq “ 1u acts on X “ Rn by left
matrix multiplication :
τA pxq “ A ¨ x
(matrix product pn ˆ nq ¨ pn ˆ 1q “ pn ˆ 1q)
if we regard vectors in Rn as n ˆ 1 column matrices. There is at least one trivial orbit
G¨0 “ t0u because the τA are linear operators on Rn . Prove that this action is transitive
on the complementary set X “ Rn z-t0u so there are just two orbits.
Hints : First, explain why it suffices to show that the unit vector e1 “ colp1, 0, . . . , 0q
can be moved to any other nonzero vector x by suitably chosen A P SL. Remember that
A ¨ e1 “(the first columnn of A) for any A P Mn pRq.
Solution :
If x, y ‰ 0 in Rn are given and we can find A, B P SLn pRq such that Ae1 “ x, Be1 “ y
then e1 “ Ax and y “ Be1 “ BA´1 x with detpBA´1 q “ detpBq{detpAq “ 1. We are
reduced to showing : @x ‰ 0, there is A such that Ae1 “ x and detpAq “ 1.
If LA x “ Ax viewing x as an n ˆ 1 column vector,
¨ ˛
1
˚ 0 ‹
˚ ‹
‹
Ae1 “ A ¨ ˚
˚ . ‹ “ Column 1
˝ . ‚
0
4
¨
˚
˚
of A “ ˚
˚
˝
a11
.
.
.
an1
˛
‹
‹
‹ for any matrix A “ Mn pRq. Thus if x is given ‰ 0, to get A ¨ e1 “ x,
‹
‚
¨
˛
x1
˚ . ‹
˚
‹
‹ ; but the remaining entries must be filled in so that A is
.
we need Col1 pAq “ ˚
˚
‹
˝ . ‚
xn
nonsingular and detpAq “ `1. The desired A is not unique. If x1 ‰ 0 we can take
¨
˛
x1
0 0 . 0
˚ x2 1{x1 . .
‹
˚
‹
‹
.
0
1
.
.
A“˚
˚
‹
˝ .
.
. . . ‚
xn
0
. . 1
with detpAq “ 1 because
ˆ
det
B1 0
˚ B2
˙
“ getpB1 q ¨ getpB2 q
for any matrix with block upper triangular or lower triangular form.
0
.
` . ˘
. If P is the matrix
If x1 “ 0 there must be a first entry xk ‰ 0, so x “
xk
.
xn
¨
˛
0 . . 1
.
0
˚ . 1 . . . . .
‹
˚
‹
˚ . . . . . . . . ‹
˚
‹
˚ . . 1 . . . . . ‹
P“˚
‹
˚ 1 . . 0 . . .
‹
˚ . . . . 1 . .
‹
˚
‹
˝
‚
. . . . . .
. . . . . 1
Where the ”1” outside the diagonal is at the column k and row k. (the identity matrix with
its 1; 1 at k, k entires displaces as shown), then Lp pe1 q “ ek , Lp pek q “ e1 and Lp pe j q “ e j ,
for all j ‰ 1, k. Furthermore, detpPq
řn“ ´1 since P is Inˆn with Row 1 Ø Row k.
Then Lp pe1 q “ ek , y “ Lp pxq “ Lp p i“1 xi ei q “ xk e1 ` . . . has nonzero first entry, so that
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the matrix
¨
˚
˚
˚
˚
˚
P“˚
˚
˚
˚
˝
.
.
1
.
.
.
.
.
y1
0
y2 ´1{y1
.
.
.
.
.
.
.
.
.
.
0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0
˛
‹
.
‹
. . ‹
‹
. . ‹
‹
.
‹
‹
.
‹
‚
.
1
which has detpAq “ ´1 because of the - sign on 1{y1 , yields the result Ae1 “ y “ Px
(since y1 ‰ 0). But P2 “ I, P “ P´1 and P´1 Ae1 “ x as desired.
(We have detpP´1 Aq “ detpPAq “ detpPAq “ detpPq ¨ detpAq “ p´1q2 “ A owing to the
use of ´1{y1 rather that 1{y1 in defining A.
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