Download Physics 129B, Winter 2010 Problem Set 1 Solution

Physics 129B, Winter 2010
Problem Set 1 Solution
Chan Y. Park
February 8, 2010
Problem 1
Show that any finite groups of orders 2 and 3 are isomorphic to the cyclic groups C2 and C3 .
(Hint: What are possible multiplication tables?)
Solution:
Consider a group of order 2. One of the element should be the identity, e. If we call the other
element as a (6= e), then it has an inverse, a−1 , according to the definition of a group. The group
product of a with itself, a ∗ a = a2 , should be either e or a, as there is no other element.
If a2 = a, then by multiplying a−1 on both sides of the equality, we get
a−1 ∗ a2 = a−1 ∗ a
(1)
a=e
(2)
which is contradictory to the assumption that a =
6 e. Therefore a2 = e. This completes the
multiplication table of the group of order 2, which is (3).
∗
e
a
e
e
a
a
a
e
(3)
This should be the multiplication table for all groups of order 2, because we made no assumption
for the group except that it is of order 2. So all groups of order 2, one of which is C2 , are isomorphic
to one another.
Next, consider a group of order 3, which has three elements e, a, and b. We can see that
a ∗ b = e; otherwise either a ∗ b = a or a ∗ b = b will give us b = e or a = e, respectively,
contrary to the assumption that the group is of order 3. With the fact that every row or column
of a group multiplication table has to contain all of the group elements, we can now complete the
multiplication table of a group of order 3, which is (4).
∗
e
a
b
e
e
a
b
a
a
b
e
1
b
b
e
a
(4)
Again, this applies for every group of order 3, so all groups of order 3, one of which is C3 , are
isomorphic to one another.
Problem 2
Consider the following 6 functions,
f1 (x) = x, f2 (x) =
f5 (x) =
1
1
, f3 (x) = 1 − x, f4 (x) =
x
1−x
x−1
x
, f6 (x) =
.
x
x−1
Define a product of functions fi and fj by
(fi · fj )(x) = fi (fj (x)).
Show that these 6 functions make a group under this product rule. Compute the group multiplication table. To which group, is it isomorphic?
Solution:
We can check if the binary product gives a well-defined group structure by building a multiplication
table.
One fact that is useful in the calculation of the multiplication table is that the functions can
be represented as 2 × 2 matrices and the group product of two functions corresponds to the matrix
product of corresponding two matrices, which can be easily checked by direct calculation.
ai x + bi
fi (x) =
ci x + di
(ai aj + bi cj ) x + (ai bj + bi dj )
(fi · fj ) (x) =
(ci aj + di cj ) x + (ci bj + di dj )
⇔
⇔
ai bi
Mi =
ci di
ai aj + bi cj
Mi Mj =
ci aj + di cj
(5)
ai bj + bi dj
ci bj + di dj
(6)
The multiplication table of the given binary product over {fi } is shown in (7).
·
f1
f4
f5
f2
f3
f6
f1
f1
f4
f5
f2
f3
f6
f4
f4
f5
f1
f3
f6
f2
f5
f5
f1
f4
f6
f2
f3
f2
f2
f6
f3
f1
f5
f4
f3
f3
f2
f6
f4
f1
f5
f6
f6
f3
f2
f5
f4
f1
(7)
From the table, we can see that the set {fi } with the binary product · satisfies all the axioms
of the definitions of a group, so h{fi } , ·i is indeed a group.
There are only two finite groups of order 6 – C6 , the cyclic group of order 6, and S3 , the group
of the permutations of three letters. As the given h{fi } , ·i is clearly not cyclic, it is isomorphic to
S3 . The table (7) is arranged in a way to clearly show the structure of S3 and its subgroups. f1 is
the identity, {f1 , f4 , f5 } is a subgroup that is isomorphic to C3 , and {f1 , f2 }, {f1 , f3 }, {f1 , f6 } are
subgroups that are isomorphic to C2 .
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Problem 3
Show that the cyclic group C3 is an invariant subgroup of the dihedral group of an equilateral
triangle.
Solution 1:
D6 , the dihedral group of an equilateral triangle, is a finite group of order 6 that is not cyclic.
So we expect it to be isomorphic to S3 , and the isomorphism is evident when we assign a letter
(a, b, or c) to each vertex of the equilateral triangle. Then a rotation by 2π/3 corresponds to a
permutation of order 3, e.g. (abc), and a reflection with respect to a vertex, e.g. a, corresponds to
a permutation of order 2, e.g. (bc).
Then (7) is also the multiplication table of D6 . We can rewrite it as (8), where y is an element
of order 3, x is an element of order 2, and (xy)2 = e is the relation between the two elements.
∗
e
y
y2
x
xy
xy 2
e
e
y
y2
x
xy
xy 2
y
y
y2
e
xy
xy 2
x
y2
y2
e
y
xy 2
x
xy
x
x
xy 2
xy
e
y2
y
xy
xy
x
xy 2
y
e
y2
xy 2
xy 2
xy
x
y2
y
e
(8)
From (8) we can find that the order 3 subgroup of D6 , e, y, y 2 , is isomorphic to C3 . We can
show that it is an invariant subgroup of D6 by explicit calculation.
Solution 2:
Much work can be saved if we invoke the first isomorphism theorem for groups, which states that
if we have a homomorphism ϕ : G → H between two groups G and H, then 1) ker (ϕ) E G, 2)
ϕ (G) ≤ G, and 3) ϕ (G) ∼
= G/ ker (ϕ).
For D6 = hx, y | x2 = y 3 = (xy)2 = ei, we have yx = xy 2 so every element
P of D6 , which is in
a
b
a
b
1
1
2
2
a form of x y x y · · · where ai ∈ Z2 and bi ∈ Z3 , can be rewritten as x i ai y c , where c ∈ Z3
(think of pushing all x’s to the left).
There is a homomorphism ϕ : D6 → C2 , ϕ(xa y b ) = xa where a ∈ Z2 and x0 ≡ e. The kernel
of this homomorphism is C3 = hyi, because every word that consists only of y’s is mapped to the
identity of C2 . Then from the first isomorphism theorem, we can see ker (ϕ) = C3 / D6 .
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Problem 4
If H1 and H2 are groups, then we can define a product group G = H1 ⊗ H2 as follows. G consists
of pairs (h1 , h2 ), where h1 ∈ H1 and h2 ∈ H2 , and their products are defined by (h1 , h2 ) · (h01 , h02 ) =
(h1 · h01 , h2 · h02 ). Show that H̃1 = H1 ⊗ {e} is an invariant subgroup and that G/H̃1 ' H2 .
Solution 1:
For an element g = (h1 , h2 ) ∈ G,
g H̃1 g −1 = h1 H1 h−1
⊗ h2 {e2 } h−1
= H1 ⊗ {e2 } = H̃1 ,
1
2
(9)
where ei is the identity of Hi . Therefore H̃1 is an invariant subgroup of G, and we can define a
quotient group G/H̃1 .
Now we claim that the isomorphism between G/H̃1 and H2 is given by g H̃1 = (h1 , h2 ) H̃1 ∼
= h2 .
First, we need to see if the mapping is bijective. It is easy to see that the mapping is well-defined
and surjective, so let’s check if it is injective. For g = (h1 , h2 ), g 0 = (h01 , h02 ) ∈ G,
g H̃1 = (h1 H1 ) ⊗ (h2 {e2 }) = H1 ⊗ {h2 } ,
g H̃1 = h01 H1 ⊗ h02 {e2 } = H1 ⊗ h02 .
0
(10)
(11)
So if g H̃1 6= g 0 H̃1 , then h2 6= h02 , therefore the mapping is injective.
Next, let’s prove that the mapping is a group homomorphism – that is, it preserves the binary
product as it maps G/H̃1 to H2 . For g = (h1 , h2 ), g 0 = (h01 , h02 ) ∈ G,
g H̃1 · g 0 H̃1 = g · g 0 H̃1
(12)
= (h1 , h2 ) · h01 , h02 H̃1
(13)
0
0
(14)
= h1 · h1 , h2 · h2 H̃1
0
0
= h1 · h1 , h2 · h2 (H1 ⊗ {e2 })
(15)
0
0
= h1 · h1 · H1 ⊗ h2 · h2 · e2
(16)
0
= H1 ⊗ h2 · h2
(17)
0
∼
(18)
= h2 · h2 .
Therefore the mapping defined above is a bijective group homomorphism, i.e. a group isomorphism.
Solution 2:
Again the first isomorphism theorem can be a big help here. Define a homomorphism ϕ : G → H2 ,
ϕ ((h1 , h2 )) = h2 . It is evident that ker (ϕ) = H1 ⊗ {e} = H̃1 and ϕ (G) = H2 . Therefore, by the
first isomorphism theorem, ker (ϕ) = H̃1 E G and ϕ (G) = H2 ∼
= G/H̃1 .
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