MATH 433
April 1, 2015
Exam 2: Solutions
Problem 1 (20 pts.)
disjoint cycles.
Write the permutation π = (1 3 5)(2 4)(1 2 3 4 5 6) as a product of
Solution: π = (1 4)(2 5 6 3).
Keeping in mind that the composition is evaluated from the right to the left, we find that π(1) = 4 and
π(4) = 1. Further, π(2) = 5, π(5) = 6, π(6) = 3, and π(3) = 2. Thus π = (1 4)(2 5 6 3).
Problem 2 (20 pts.) Consider a relation ∼ on the symmetric group S(n) defined as follows.
For any π, σ ∈ S(n) we let π ∼ σ if and only if π = τ στ −1 for some permutation τ ∈ S(n). Show
that ∼ is an equivalence relation.
We have to show that the relation ∼ is reflexive, symmetric, and transitive.
Reflexivity. π ∼ π for all π ∈ S(n) since π = τ πτ −1 holds for τ = id (as well as for τ = π).
Symmetry. Assume π ∼ σ, that is, π = τ στ −1 for some τ ∈ S(n). Then
σ = τ −1 πτ = τ −1 π(τ −1 )−1 = τ0 πτ0−1 ,
where τ0 = τ −1 ∈ S(n). Hence σ ∼ π.
Transitivity. Assume π ∼ σ and σ ∼ ρ, that is, π = τ1 στ1−1 and σ = τ2 ρτ2−1 for some τ1 , τ2 ∈ S(n).
Then
π = τ1 (τ2 ρτ2−1 )τ1−1 = (τ1 τ2 )ρ(τ2−1 τ1−1 ) = (τ1 τ2 )ρ(τ1 τ2 )−1 = τ ρτ −1 ,
where τ = τ1 τ2 ∈ S(n). Hence π ∼ ρ.
Problem 3 (20 pts.)
Which numbers greater than 12 are orders of permutations in the
symmetric group S(9)? Provide an example for every order.
Solution: (1 2 3 4 5 6 7)(8 9) has order 14, (1 2 3 4 5)(6 7 8) has order 15, and (1 2 3 4 5)(6 7 8 9)
has order 20.
The order o(π) of a permutation π is the least common multiple of l1 , l2 , . . . , lk , the lengths of cycles in
the disjoint cycle decomposition of π. If π is a cycle then o(π) ≤ 9. If all lengths li divide 12 then o(π) ≤ 12.
Therefore it is no loss to assume that π is the product of at least two nontrivial disjoint cycles and that the
length of the longest cycle is either 5 or more than 6. This leaves only a handful of cycle shapes to consider.
Three of them provide examples of permutations of order greater than 12.
Problem 4 (20 pts.) Consider a binary operation ∗ on the semi-closed interval X = [0, 1)
defined as follows. For any x, y ∈ X we let
x+y
if x + y < 1,
x∗y =
x + y − 1 if x + y ≥ 1.
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Is (X, ∗) a semigroup? Is (X, ∗) a group? Explain.
Solution: Yes, (X, ∗) a semigroup and a group.
We have to check four axioms.
Closure. If x, y ∈ X then 0 ≤ x + y < 2. In the case x + y < 1, we obtain 0 ≤ x + y < 1. In the case
x + y ≥ 1, we obtain 0 ≤ x + y − 1 < 1. In either case, x ∗ y ∈ X.
Associativity. Let x, y, z ∈ X. Note that (x ∗ y) ∗ z equals x ∗ y + z or x ∗ y + z − 1. Also, x ∗ y equals
x + y or x + y − 1. Therefore (x ∗ y) ∗ z is one of the numbers x + y + z, x + y + z − 1, and x + y + z − 2.
Similarly, x ∗ (y ∗ z) is one of the same three numbers. It is easy to observe that at most one of the three
numbers can belong to X. Since (x ∗ y) ∗ z and x ∗ (y ∗ z) are both in X (due to the closure axiom, which has
already been verified), we conclude that one of the numbers x + y + z, x + y + z − 1, and x + y + z − 2 does
belong to X and, moreover, it coincides with both (x ∗ y) ∗ z and x ∗ (y ∗ z). Thus (x ∗ y) ∗ z = x ∗ (y ∗ z).
Existence of the identity element. It is easy to see that x ∗ 0 = 0 ∗ x = x for all x ∈ X. Hence 0
is the identity element of the operation ∗.
Existence of the inverse. We observe that x ∗ (1 − x) = (1 − x) ∗ x = 0 for all x ∈ (0, 1). Hence
1 − x is the inverse of x for 0 < x < 1. Besides, the inverse of 0 is clearly 0 itself.
√
Problem 5 (20 pts.) Let M be the set of all numbers of the form p + q 5, where p, q ∈ Q.
Under the operations of addition and multiplication, does this set form a ring? Does M form a field?
Explain.
Solution: Yes, the set M forms a ring and a field.
√
First we show
that
the
set
M
is
closed
under
addition
and
multiplication.
Let
r,
r̃
∈
M
.
Then
r
=
p+q
5
√
and r̃ = p̃ + q̃ 5 for some p, q, p̃, q̃ ∈ Q. We obtain
√
r + r̃ = (p + p̃) + (q + q̃) 5,
√
rr̃ = (pp̃ + 5q q̃) + (pq̃ + p̃q) 5,
which implies that r + r̃, rr̃ ∈ M .
√
Further, the set√M contains the additive identity element 0√= 0 + 0 5 and the multiplicative identity
element 1 = 1 + 0 √5. Besides, with every element r = p + q 5, the set M contains its additive inverse
−r = (−p) + (−q) 5. Associativity and commutativity of addition, associativity and commutativity of
multiplication, and the distributive law need not be verified because they hold for all real numbers. Thus
M is a commutative ring with 1.
To prove that M is a field, it remains to
√ show that for any non-zero number r ∈ M , its multiplicative
−1
inverse
r is also in M . We have r =√p + q 5, where p, q ∈ Q are not both equal to zero. Since the number
√
5 is irrational, it follows that p − q 5 6= 0. Then
√
√
p−q 5
p−q 5
p
−q √
1
−1
√ =
√
√ = 2
= 2
+ 2
r =
5,
2
2
p − 5q
p − 5q
p − 5q 2
p+q 5
(p + q 5)(p − q 5)
which implies that r −1 ∈ M .
Bonus Problem 6 (15 pts.) Construct an acceptor automaton over the alphabet {a, b} that
accepts only those words that contain an even number of a’s and an odd number of b’s.
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Solution: One of possible solutions is an automaton with four states 0, 1, 2, 3 and a state
transition function given by the following table:
0
1
2
3
a
2
3
0
1
b
1
0
3
2
The initial state is 0. The only accepting state is 1.
The above automaton memorizes some information about the previous input by changing its state.
Namely, it is in the state 0 if the previous input contains an even number of a’s and an even number of b’s,
it is in the state 1 if the previous input contains an even number of a’s and an odd number of b’s, it is in
the state 2 if the previous input contains an odd number of a’s and an even number of b’s, and it is in the
state 3 if the previous input contains an odd number of a’s and an odd number of b’s.
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