Dr. Marques Sophie
Office 519
Algebra 1
Fall Semester 2014
marques@cims.nyu.edu
Problem Set 8
Exercise 1 :
The units U9 “ trks : gcdpk, nq “ 1u of the set Z{9Z form a group under the multiplication p¨q.
1. Identify the elements in U9 and make a list shows
(a) The element in the subgroup Hx “ă x ą,
(b) The order opxq.
for each x P U9 .
2. Identify the cyclic generators of U9 , if any exist,
3. If n ě 3 prove that |Un | is always even,
4. By examining Un for small n ě 3 provide a counterexample showing that Un “ă
a ą does not always implies that Un is generated by p´1q ¨ a.
Hint : In 3., if n ě 3, then r´1s ‰ r1s in Un .
Solution :
1. U9 “ tr1s, r2s, r4s, r5s, r7s, r8su with r8s “ ´r1s. Generated subgroups are listed
below :
x opxq ă x ą
1 1
1
2 6
1, 2, 4, 8, 16 ” 7, 14 ” 5, 10 ” 1 ˚
4 3
1, 4, 16 ” 7, 28 ” 1
5 6
1, 5, 25 ” 7, 35 ” ´1, ´5, ´25 ” 2, 10 ” 1 ˚
7 3
1, 7, 49 ” 4, 28 ” 1
8 2
1, 8, 64 ” 1
˚ “ cyclic generator.
2. x is a generator if and only if opxq “ |U9 | “ 6, so x “ r2s and x “ r5s “ r´4s are
the cyclic generators of U9 .
3. The map J : x ÞÑ ´x “ r´1s ¨ x is a bijection on UN with J2 “ Id (so J “ J´1 ).
If n ě 3 then r´1s ‰ r1s in Un and Jpxq ‰ x all x P Un (because if p´1q ¨ x “ x
for some x, multiplying by x´1 given r´1s “ r´1s ¨ xx´1 “ x ¨ x´1 “ r1s and
1 ” ´1pmod nq ” n “ 2.]
If |Un | were odd then would be a partition of Un into pairs tx1 , Jpx1 qu, . . . , txr , Jpxr qu,
with Jpxi q ‰ xi , plus are leftover element xr`1 . But then Jpxr`1 q could only “ xr`1 , a
contradiction because Jpxr`1 q “ ´xr`1 ‰ xr`1 .
1
4. Trying n “ 3, 4, 5 we find that U5 “ t1, 2, 3, 4u has cyclic subgroups such that
|U5 | “ 4. r1s is a cyclic generator but r´1s ¨ r2s “ r3s is not.
x
1
2
´2 ” 3
´1 ” 4
opxq
1
4
3
2
ăxą
1
1, 2, 4, 8 ” 3, 16 ” 1 ˚
1, 3, 9 ” 4
1, 4, 16 ” 1
r1s is a cyclic generator but r´1s ¨ r2s “ r3s is not.
Exercise 2 :
The commutator subgroup rG, Gs of a group G is the subgroup rG, Gs “ă S ą generated
by the set of all commutators
rx, ys “ xyx´1 y´1 , pall x, y P Gq
1. Prove that rG, Gs is a normal subgroup in G ;
2. Prove that the quotient group G{rG, Gs is commutative.
Hints : For normality of a subgroup H you must show gHg´1 Ď H for every g P G.
What does a conjugation x ÞÑ gxg´1 do to a commutator ?
Solution :
1. A typical element in S is s “ xyx´1 y´1 for x, y P G. Then
gsg´1 “ gxyx´1 y´1 g´1 “ gxg´1 ¨ gyg´1 ¨ gpx´1 qg´1 ¨ gpy´1 qg´1
and since x ÞÑ gxg´1 is an isomorphism, we have gxg´1 “ pgxg´1 q´1 and likewise
for y. Thus gsg´1 “ gxg´1 ¨ gyg´1 ¨ pgxg´1 q´1 ¨ pgyg´1 q “ rgxg´1 , gyg´1 s which is
just another commutator in S, so gSg´1 “ S for any g P G. Also, note that S´1 “ S
because the inverse of a generator rx, ys is
rx, ys´1 “ pxyx´1 y´1 q´1 “ py´1 q´1 px´1 q´1 y´1 x´1 “ yxy´1 x´1 “ ry, xs
(another commutator in S). Thus S Y S´1 “ S. Now, rG, Gs “ă S ą implies a typical
element u P rG, Gs is a finite-length ”word” u “ a1 . . . ar with ai P S Y S´1 “ S and
r ă 8. Then gug´1 P S Ďă S ą“ rG, Gs. Thus, for all g P G, we have proved
grG, Gsg´1 “ g ă S ą g´1 Ďă S ą“ rG, Gs ;
2. Write x̄ for πpxq P G{rG, Gs if x P G. Then xyx´1 y´1 P kerpπq “ rG, Gs, if x̄ “
πpxq, ȳ “ πpyq P Ḡ “ G{rG, Gs, we get x̄´1 “ πpxq´1 “ πpx´1 q “ x´1 , ȳ “ y´1
and
x̄ ȳpx̄q´1 p ȳq´1 “ πpxqπpyqπpx´1 qπpy´1 q
“ πpxyx´1 y´1 q “ πprx, ysq
But π bills all commutators since kerpπq “ rG, Gs, so x̄ ȳpx̄q´1 p ȳq´1 “ ē (the identity
in Ḡ). Or x̄ ȳ “ ȳx̄, @x̄, ȳ P Ḡ. Ḡ is abelian as claimed.
2
Exercise 3 :
1. If x, y are element in a group G such that they commute that is xy “ yx and
have finite order that is opxq “ m and opyq “ n, prove that the order opxyq of
their product is always a divisor of the least common multiple lcmpopxq, opyqq “
lcmpm, nq.
2. If x “ pi1 , . . . , im q and y “ pj1 , . . . , jn q are disjoints cycles in the permutation group
SN , explain why the order of their product xy is always equal to the least common
multiple opxyq “ lcmpm, nq “ lcmpopxq, opyqq.
Solution :
1. Let x, y in a group G such that they commute that is xy “ yx and have finite order
that is opxq “ m and opyq “ n, then m|lcmpm, nq and n|lcmpm, nq with lcmpm, nq “
m ¨ a “ n ¨ b. Thus pxyqlcmpm,nq “ xlcmpm,nq ylcmpm,nq “ pxm qa pyn qb “ ea eb “ e (because
G is abelian). Then opxyq divides lcmpm, nq.
2. The cycles have order = (length of cycle). So opxq “ m and opyq “ n. Since, they
have disjoint supports ti1 , . . . , im u and tj1 , . . . , jn u they act on different parts of the
space r1, Ns. Thus pxyqk “ xk yk (disjoint cycles in SN commute) and the action of
pxyqk on u P r1, Ns is :
pxyqk puq “ x$k yk puq “ xk pyk puqq
i f u R supppxq Y supppyq
& u
k
y
puq
i
f u P supppyq psince yk puq is also in supppyq.q
“
% k
x puq i f u P supppxq psince this implies yk puq “ uq
Now pxyqk “ e in Sn ô pxyqk puq “ Idpuq “ u @u P r1, ns
ô xk puq “ u @u P supppxq, yk puq “ u, @u P supppyq (and xk puq “ yk puq “ u for u
outside supppxq Y supppyq by defn of ”support”)
ô xk “ Id and yk “ Id on all of r1, ns (x and y act trivially if and only if of their
supports)
ô xk “ e and yk “ e in Sn .
In particular, if k “ opxyq, we have xk “ yk “ e, so opxq divides k and opyq divides k.
lcmpopxq, opyqq divides k “ opxyq. But in 1. we showed that opxyq|lcmpm, nq. For any
a, b ‰ 0 in Z, a|b and b|a ñ b “ ˘a so opxyq “ lcmpopxq, opyqq.
Exercise 4 :
The group G of matrices
ˆ
A“
1 b
0 1
˙
with b P C and detpAq ‰ 0
is a subgroup of the full ”upper triangular group” M of matrices
ˆ
˙
a b
B“
with a, b, c, d P C and detpBq “ ab ‰ 0
0 d
3
1. Prove that G is normal in M. Is it normal subgroup in the full linear group
GL2 pCq ?
2. Prove that the quotient M{G is abelian.
Solution :
ˆ
˙
1 ´t
1. If A “
then A “ A “
and G is a group because
0 1
ˆ
˙ˆ
˙ ˆ
˙
1 t
1 s
1 s`t
“
PG
0 1
0 1
0 1
ˆ
˙
a b
Then if B “
we have
0 d
ˆ
˙ˆ
˙ˆ
˙ ˆ
˙
a b
1 t
1{a ´b{a
1 pa{dq ¨ t
´1
BAB “
“
0 d
0 1
0
1{a
0 1
1 t
0 1
˙
ˆ
´1
which is again in G. Thus G is normal in M.
ˆ
˙
1 1
But G is not normal in GL2 pCq. For instance, if C “
, we have
1 0
ˆ
˙ˆ
˙ˆ
˙ ˆ
˙
1 1
1 t
1
0
a´b
b
´1
CBC “
“
1 0
0 1
´1 1
a´b´d b`d
which is not in G for all C P GL2 pCq.
2. If A, A1 P M we must show pAA1 q ¨ G “ pAGq ¨ pA1 Gq “ pA1 Gq ¨ pAGq “ pA1 Aq ¨ G.
This is true if and only if pA1 Aq´1 pAA1 q ¨ G “ G (the identity coset in M{G) ô
1
´1
1 ´1 1
´1
1 ´1
pA1 Aq´1 pAA
ˆ q “ A˙ pA q A
ˆ A1 “ p1 the
˙ commutator rA , pAˆq s) is in
˙ G.
a b
a
b
d
´b
Let A “
, A1 “
, so A´1 “ 1{detpAq
, pA1 q´1 “
1
0
d
0
d
0
a
ˆ 1
˙
d ´b1
1
1{detpA q
. Then
0 a1
A´1 pA1 q´1 AA1
ˆ
˙ˆ 1
˙ˆ
˙ˆ 1 1 ˙
d ´b
d ´b1
a b
a b
“ 1{pdetpAqdetpA qq
1
0 a
0 ˙d
0 d1
ˆ 1
˙ ˆ0 1a 1
1
1
1
dd ´db ´ ba
aa ab ` bd
“ 1{pada1 d1 q
1
1
ˆ 0 1 aa
˙ ˆ 0 dd˙
1
aa dd ˚
1 ˚˚
“ 1{paa1 dd1 q
“
1
1
0
aa dd
0 1
1
The nature of the terms p˚q, p˚˚q does not matter ; it is evident that the product is in
the identity coset I ¨ G “ G (ie. is congruent to I (mod G)). Thus M{G is an abelian
group.
4