Physics 5040
Spring 2009
Problem Set 8 Solutions
1. (a) For the set (Z, −) we have 0 as the identity element, so each integer is its
own inverse. However, this is not a group because n − 0 6= 0 − n, and also
note that subtraction is not associative since m − (n − q) 6= (m − n) − q.
(b) The set G = (Q − {0}, ÷) is not a group. A potential inverse is 1, but
1 ÷ x 6= x ÷ 1. Furthermore, division is not associative.
(c) This is just the cyclic group of order 7, C7 . The identity is a0 and the
inverses are (a0 )−1 = a0 and (ai )−1 = a7−i for 1 ≤ i ≤ 6.
(d) This forms a group with identity element 1 and inverse given by (2m 3n )−1 =
2−m 3−n .
2. Given f ∈ F [R], define 0 ∈ F [R] by 0(x) = 0 and −f ∈ F [R] by (−f )(x) =
−f (x). Then we make F [R] into a group by defining the identity element to
be e = 0 ∈ F [R] and for every f ∈ F [R] we let f −1 = −f ∈ F [R]. To show
that F [R] satisfies the group axioms, we have for every f, g, h ∈ F [R]
(i)
(f + g)(x) = f (x) + g(x)
so that f + g ∈ F [R].
(ii)
[(f + g) + h](x) = (f + g)(x) + h(x) = f (x) + g(x) + h(x)
= f (x) + (g + h)(x) = [f + (g + h)](x)
so that (f + g) + h = f + (g + h).
(iii) Clearly
(f + 0)(x) = f (x) + 0 = f (x) = 0 + f (x) = (0 + f )(x)
so that f + 0 = 0 + f = f .
(iv)
[f + (−f )](x) = f (x) − f (x) = 0 = −f (x) + f (x) = [(−f ) + f ](x)
so that f + (−f ) = (−f ) + f = 0.
3. (a) If A, B are groups, and if (a1 , b1 ), (a2 , b2 ) ∈ A × B with (a1 , b1 )(a2 , b2 ) :=
(a1 a2 , b1 b2 ) ∈ A × B, then this satisfies the closure requirement for a
group. (Since a1 a2 ∈ A and b1 b2 ∈ B.) As to associativity, we have
[(a1 , b1 )(a2 , b2 )](a3 , b3 ) = (a1 a2 , b1 b2 )(a3 , b3 ) = (a1 a2 a3 , b1 b2 b3 )
= (a1 , b1 )(a2 a3 , b2 b3 ) = (a1 , b1 )[(a2 , b2 )(a3 , b3 )] .
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Since A, B are groups, there are identity elements eA ∈ A and eB ∈ B.
Then
(eA , eB )(a, b) = (eA a, eB b) = (a, b) = (aeA , beB ) = (a, b)(eA , eB )
so that the identity in A × B is just (eA , eB ). As to the inverse, again we
use the fact that A and B are groups so every a ∈ A and b ∈ B has an
inverse. Then
(a−1 , b−1 )(a, b) = (a−1 a, b−1 b) = (eA , eB ) = (aa−1 , bb−1 ) = (a, b)(a−1 , b−1 )
and hence (a, b)−1 = (a−1 , b−1 ).
(b) If G is a group, consider the set D = {(x, x) ∈ G × G}. With group
operations defined as in part (a), we have for the group axioms:
(i)
(x, x)(y, y) = (xy, xy) ∈ D
(since xy ∈ G) so D is closed.
(ii)
[(x, x)(y, y)](z, z) = (xy, xy)(z, z) = (xyz, xyz) = (x, x)(yz, yz)
= (x, x)[(y, y)(z, z)]
so we have associativity.
(iii) If e is the identity in G, then (e, e) ∈ D and
(e, e)(x, x) = (ex, ex) = (x, x) = (xe, xe) = (x, x)(e, e)
so that (e, e) is the identity in D.
(iv) Every x ∈ G has an inverse x−1 ∈ G so that (x−1 , x−1 ) ∈ D and
(x, x)(x−1 , x−1 ) = (xx−1 , xx−1 ) = (e, e) = (x−1 x, x−1 x)
= (x−1 , x−1 )(x, x)
so that (x, x)−1 = (x−1 , x−1 ) ∈ D.
4. (a)
(123)(45)(16789)(15) = (123)(45)(156789) = (123)(1456789)
= (145678923) .
(b)
(12)(123)(12) = (12)(13) = (132) .
5. (a) Let ψ : A → B and φ : B → C be homomorphisms. Then φ ◦ ψ : A → C
is a homomorphism because
(φ ◦ ψ)(a1 a2 ) = φ(ψ(a1 a2 )) = φ(ψ(a1 )ψ(a2 )) = φ(ψ(a1 ))φ(ψ(a2 ))
= (φ ◦ ψ)(a1 )(φ ◦ ψ)(a2 ) .
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(b) Let G denote the set of all groups, and let Φ be the set of all group
isomorphisms. We say that G1 , G2 ∈ G are equivalent if G2 = φ(G1 ) for
some φ ∈ Φ, and we write this as G1 ∼ G2 . To show that this defines an
equivalence relation on G , we have
(i) Given G ∈ G , let id be the identity map on G. This is clearly an
isomorphism (since id(ab) = ab = id(a)id(b)) so that id ∈ Φ. Then
G = id(G) so that G ∼ G.
(ii) If G1 ∼ G2 , then φ(G1 ) = G2 for some φ ∈ Φ. But φ−1 ∈ Φ so we
have G1 = φ−1 (G2 ) so that G2 ∼ G1 .
(iii) If G1 ∼ G2 and G2 ∼ G3 , then G2 = φ(G1 ) and G3 = ψ(G2 ) for
some φ, ψ ∈ Φ. By part (a) we know that ψ ◦ φ is a homomorphism
and hence also an isomorphism, and
(ψ ◦ φ)(G1 ) = ψ(φ(G1 )) = ψ(G2 ) = G3
so that G1 ∼ G3 (since ψ ◦ φ ∈ Φ).
6. (a) Let G = (R − {0}, ·) and let ϕ : G → G be defined by ϕ(x) = x2 . We first
note that the identity on G is 1 ∈ R, and that ϕ(1) = 1 as it must for a
homomorphism. Then we also see that
ϕ(xy) = (xy)2 = x2 y 2 = ϕ(x)ϕ(y)
and hence ϕ is a homomorphism. If ϕ(x) = x2 = 1, then we must have
x = ±1 so that Ker ϕ = {1, −1}. Note this also shows that ϕ is not an
isomorphism.
(b) Let G = (R, +) and let ϕ : G → G be defined by ϕ(x) = 1 + x. Since
ϕ(x + y) = 1 + (x + y) 6= (1 + x) + (1 + y)
we see that ϕ is not a homomorphism. We could also observe that ϕ(0) = 1
which is not the identity on G (which is 0).
(c) Let G = (R, +) and let ϕ : G → G be defined by ϕ(x) = kx for some
fixed k ∈ R. (Assume that k 6= 0 or ϕ is just the zero map.) Now we have
ϕ(0) = 0 as it must for a homomorphism. Furthermore, we have
ϕ(x + y) = k(x + y) = kx + ky = ϕ(x) + ϕ(y)
so that ϕ is a homomorphism. Since ϕ(x) = kx = 0 implies that x = 0,
we see that Ker ϕ = {0} and ϕ is an isomorphism.
7. (a) Let ϕ : G → G′ be a homomorphism with kernel Kϕ . We first show that
Kϕ is a subgroup. If k1 , k2 ∈ Kϕ , then ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) = e′ e′ = e′
so that k1 k2 ∈ Kϕ . Next, if k ∈ Kϕ , then ϕ(k) = ϕ(ke) = ϕ(k)ϕ(e)
implies that ϕ(e) = e′ so that e ∈ Kϕ . Finally, for any k ∈ Kϕ we
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have e′ = ϕ(e) = ϕ(kk −1 ) = ϕ(k)ϕ(k −1 ) = e′ ϕ(k −1 ) = ϕ(k −1 ) so that
k −1 ∈ Kϕ . This shows that Kϕ is a subgroup of G.
Now, to show that Kϕ is invariant (i.e., normal), simply note that if g ∈ G
and k ∈ Kϕ , then
ϕ(gkg −1 ) = ϕ(g)ϕ(k)ϕ(g −1 ) = ϕ(g)e′ ϕ(g −1 ) = ϕ(gg −1 ) = ϕ(e) = e′
so that gkg −1 ∈ Kϕ . Hence Kϕ is a normal subgroup of G.
(b) Define the center of G by Z = {z ∈ G : zg = gz for all g ∈ G}. To show
that this is a subgroup, we note that if z1 , z2 ∈ Z, then for any g ∈ G
we have z1 z2 g = z1 gz2 = gz1 z2 which proves closure; associativity follows
from associativity in G; Z contains the identity since eg = ge; and if
z ∈ Z, then zg = gz so multiplying from the right and left by z −1 yields
gz −1 = z −1 g so that z −1 ∈ Z. Therefore Z is a subgroup of G. Finally,
it is easy to see that Z is normal because gzg −1 = zgg −1 = z ∈ Z.
8. (a) The group D2 consists of the set {e, a, b, c} where a and b represent rotations by π about the two symmetry axes of the rectangle, and c represents
a rotation by π about the perpendicular axis. Clearly a2 = b2 = c2 = e.
Now consider the product ab. Labeling the corners of the rectangle for
ease of identification, we see that starting from
3
2
b
4
1
a
we first apply b to obtain
4
1
b
2
3
a
and then a to obtain
4
1
4
b
2
3
a
which has the same net effect as simply c, and hence we see that ab = c.
To evaluate ac, we apply a to the last figure above (which was just the
result of c applied alone) to obtain
4
1
b
2
3
a
which is the same as only b, so that
we easily find
e a
e e a
a a e
b b c
c c b
ac = b. Continuing in this manner
b
b
c
e
a
c
c
b .
a
e
However, you can save yourself some work if you realize that each row and
column must contain all of the group elements (this is just the rearrangement theorem), so if you find three of them, then you know the fourth.
(b) Let us write D2 = {e, a, b, c} = {g1 , g2 , g3 , g4 }. Multiplying from the
left by e doesn’t change anything, so gi = egi = gei corresponds to
the identity permutation pe = (1)(2)(3)(4) ∈ S4 . Next, multiplying by
a yields {a, e, c, b} = {g2 , g1 , g4 , g3 } so that (a1 , a2 , a3 , a4 ) = (2, 1, 4, 3)
which is the permutation pa = (12)(34) ∈ S4 . Multiplying by b we
have {b, c, e, a} = {g3 , g4 , g1 , g2 } so (b1 , b2 , b3 , b4 ) = (3, 4, 1, 2) which corresponds to the permutation (13)(24) ∈ S4 . Finally, multiplying by c
we have {c, b, a, e} = {g4 , g3 , g2 , g1 } so (c1 , c2 , c3 , c4 ) = (4, 3, 2, 1) and the
permutation is (14)(23) ∈ S4 . This shows that D2 corresponds to the
elements
{(1)(2)(3)(4), (12)(34), (13)(24), (14)(23)} ∈ S4 .
We must still show that this is a subgroup. It contains the identity, and
using the fact that disjoint cycles commute together with the fact that
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2-cycles are their own inverses, you can easily show that this is indeed
a subgroup of S4 . For example, (12)(34) · (13)(24) = (14)(23) which
corresponds to ab = c.
9. If D is reducible, then there is an invariant subspace and a similarity transformation R that takes the matrix representation of D to block triangular form
T = R−1 DR. Solving this for D we have D = RT R−1 . Then if U = SDS −1 ,
we have U = S(RT R−1)S −1 = (SR)T (SR)−1 so that T = (SR)−1 U (SR).
This shows that the transformation SR puts U into block triangular form,
and hence U is reducible.
10. Let xi be a representative element of Ci , so that Ci contains all elements
of the form gxi g −1 for all g ∈ G. Similarly, Cj contains all elements of
the form hxj h−1 . If Ci Cj contains e, then for some g, h we must have
e = (gxi g −1 )(hxj h−1 ). This is the same as g −1 h = xi g −1 hxj or xj =
−1
(g −1 h)−1 x−1
h) ∈ Ci′ . Thus, if j 6= i′ , Ci Cj can not contain e = [e] = C1 .
i (g
Probably an easier way to say the same thing is let xi ∈ Ci , xj ∈ Cj and
suppose that xi xj = e. Then xj = xi−1 ∈ Ci′ . But classes are either disjoint
or identical, so this would imply that Cj = Ci′ .
11. It might be helpful to look at this diagram:
φ
G′
G
ψ
G/Kφ
(a) Let the homomorphism φ : G → G′ have kernel Kφ , and define ψ :
G/Kφ → G′ by ψ(X) = φ(g) where X = Kφ g. To show that ψ is welldefined, suppose that Kφ g = Kφ h. Then g = kh for some k ∈ Kφ , and
hence
φ(g) = φ(kh) = φ(k)φ(h) = e′ φ(h) = φ(h)
so that ψ is well-defined. To show that ψ is a homomorphism, let X =
Kφ g ∈ G/Kφ , Y = Kφ h ∈ G/Kφ and simply calculate using the fact that
Kφ is a normal subgroup and φ is a homomorphism:
ψ(XY ) = ψ(Kφ gKφ h) = ψ(Kφ Kφ gh) = ψ(Kφ gh) = φ(gh)
= φ(g)φ(h) = ψ(X)ψ(Y ) .
To show that ψ is surjective, let g ′ ∈ G′ be arbitrary. Since φ is surjective,
there exists g ∈ G such that g ′ = φ(g). But then ψ(Kφ g) = φ(g) = g ′
so that ψ is surjective. Lastly, to show that Ker ψ = Kφ (the identity in
G/Kφ ), suppose that ψ(Kφ g) = e′ . Then φ(g) = ψ(Kφ g) = e′ so that
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g ∈ Kφ , and hence Kφ g = Kφ . Thus Ker ψ = Kφ which is the zero element (or the identity) in G/Kφ . This completes the proof that ψ is an
isomorphism of G/Kφ onto G′ .
(b) To say U (G) is degenerate means that the homomorphism ϕ : G →
U (G) has a non-trivial kernel Kϕ , which from Problem 7a we know is
a normal subgroup. But then part (a) of this problem shows that G/Kϕ
is isomorphic to U (G), which thus defines a faithful representation of
G/Kϕ .
12. From Example 16 we know that S3 has three classes C1 = {e}, C2 = {(12),
(23), (31)} and C3 = {(123), (321)}, and hence Theorem 15 tells us there
are three inequivalent irreps. Let D1 (S3 ) denote the one-dimensional identity
representation p → 1 for every p ∈ S3 . Clearly the corresponding three
characters (one for each class) are (1, 1, 1). From Example 21 we have a
second one-dimensional induced representation D2 (g) = +1 for g ∈ C1 , C3 ,
and D2 (g) = −1 for g ∈ C2 . This then gives
Pus the characters for each class
as (1, −1, 1). From equation (11) we have µ nµ 2 = nG , and since nG = 6
for S3 , the only solution to this is n1 = 1, n2 = 1 and n3 = 2, and therefore
there is a two-dimensional irrep. We need the characters χ3 (Ci ). Since the
representation of e ∈ S3 is the two-dimensional identity matrix, we know that
χ3 (C1 ) = 2. The character table up to this point is
D1
D2
D3
C1
1
1
2
3C2 2C3
1
1
.
−1
1
To fill in the remaining two entries we use the fact that the columns are
orthogonal in the sense of equation (12), which in this (real) case becomes
X
χµ (Ck )χµ (Ci ) =
µ
6
δki .
nk
If we let k = i = 2, then nk = 3 and this equation says 12 +(−1)2 +[χ3 (C2 )]2 =
2 so that we must have χ3 (C2 ) = 0. Alternatively, we could let k = 1 and
i = 2 to write 1 ·1 + 1 ·(−1)+ 2 ·χ3(C2 ) = 0 to again conclude that χ3 (C2 ) = 0.
Finally, the same argument applied to the first and third columns shows that
χ3 (C3 ) = −1, and we have
D1
D2
D3
C1
1
1
2
7
3C2 2C3
1
1
.
−1
1
0 −1