Math 415.502
Exam 2
Solutions
November 23, 2010
As can easily be expected, the solutions provided below are not the only ways to solve these
problems, and other solutions may be completely valid. If you have questions about what went
wrong (or right!) with your own solutions, please talk to me about it so that you will improve on
the final.
1. [18 points] Consider the elements σ and τ of
1 2 3 4 5 6 7
σ=
2 8 6 4 3 5 7
the symmetric group S8 :
8
,
τ = (1, 4, 3, 2)(2, 4, 7).
1
(a) Write σ and τ as the product of disjoint cycles.
(b) What is the order of τ ?
(c) Is στ an element of A8 ?
Solution: (a) It is straightforward to check that σ = (1, 2, 8)(3, 6, 5) and τ = (1, 4, 7)(2, 3).
(b) As the product of disjoint cycles, τ is the product of a 2-cycle and a 3-cycle. The order
of τ is then the least common multiple of the lengths of these cycles, which is 6.
(c) There are several ways to do this. One is to write σ and τ as products of transpositions:
σ = (1, 8)(1, 2)(3, 5)(3, 6),
τ = (1, 7)(1, 4)(2, 3).
Therefore, σ is even, while τ is odd. Therefore στ is odd. Since A8 consists of all even
permutations in S8 , we see that στ ∈
/ A8 .
2. [12 points] Consider the ring Z8 .
(a) Is Z8 an integral domain? Why or why not?
(b) Find all elements x ∈ Z8 that satisfy x2 − 1 = 0 in Z8 . What is unusual about your
answer?
Solution: (a) No, Z8 is not an integral domain. Since 2 · 4 = 0 in Z8 , we see that 2 and 4
are both zero-divisors in Z8 .
(b) We check for possible solutions: in Z8 ,
02 − 1 = 7,
12 − 1 = 0,
22 − 1 = 3,
32 − 1 = 0,
42 − 1 = 7,
52 − 1 = 0,
62 − 1 = 3,
72 − 1 = 0.
Thus there are four solutions x = 1, 3, 5, 7. The answer to second part of the question was a
bonus: what is unusual is that the quadratic equation x2 − 1 = 0 has 4 distinct solutions in
Z8 .
3. [(a) 10 points; (b) 10 points; (c) 6 points] Let
a b
G=
∈ M2 (R) a 6= 0 .
0 a
Take it as given that G is a group under matrix multiplication. Consider the function
a b
∗
φ:G→R ,
φ
= a,
0 a
(a) Show that φ is a group homomorphism (where as usual R∗ is a group under multiplication).
(b) What is the kernel of φ?
(c) Write down the isomorphism that we can deduce from the Fundamental Homomorphism
Theorem.
a b
c d
Solution: (a) Let
,
∈ G. Then
0 a
0 c
φ
a b
c d
ac ad + bc
=φ
0 a
0 c
0
c
= ac
c d
a b
.
φ
=φ
0 c
0 a
Therefore, φ is a homomorphism.
(b) By definition
Ker(φ) = {A ∈ G | φ(A) = 1},
since 1 is the identity element of R∗ . Thus we see easily that
1 b Ker(φ) =
b∈R .
0 1 (c) For a group homomorphism φ : G → G0 with kernel K, the Fundamental Homomorphism
Theorem produces a natural isomorphism,
ψ : G/K → φ[G],
such that ψ(aK) = φ(a) for all aK ∈ G/K. In our particular case, we observe easily that φ
maps onto R∗ , so we obtain an isomorphism
1 b b ∈ R → R∗ ,
ψ:G
0 1 defined by ψ
a b
Ker(φ) = a.
0 a
4. [24 points] Let G be a finite group of order m, and let H be a normal subgroup of G with
index (G : H) = k.
(a) Complete the following statement and then prove: For all a ∈ G, am =
.
(b) Prove: For all a ∈ G, we have ak ∈ H.
Solution: (a) Claim: For any a ∈ G, am = e. Proof: Let a ∈ G. Let d be the order of a.
Then by definition of order, d = |hai|. By Lagrange’s Theorem, |hai| divides |G|, and so d
divides m. Therefore, m = dc for some positive integer c, and thus
am = adc = (ad )c = ec = e,
since ad = e.
(b) Let a ∈ G. By definition (G : H) = |G/H|, and so |G/H| = k. By part (a), we see that
(aH)k = H (= identity element of G/H)
in G/H. But (aH)k = ak H, and so ak H = H. This implies that ak ∈ H, since for b ∈ G,
bH = H if and only if b ∈ H.
√
5. [20 points] Let S = {a + b 2 | a, b ∈ Q} ⊆ R.
(a) Show that S is a ring under addition and multiplication of real numbers.
(b) Show that S is a field.
√
√
Solution: (a) First we need to show that S is closed under + and ·. Let a+b 2, c+d 2 ∈ S,
with a, b, c, d ∈ Q. Then
√
√
√
(a + b 2) + (c + d 2) = (a + c) + (b + d) 2,
which is in S because a + c ∈ Q and b + d ∈ Q. Also,
√
√
√
√
√
(a + b 2)(c + d 2) = ac + bc 2 + ad 2 + 2bd = (ac + 2bd) + (bc + ad) 2,
and this is in S because ac + 2bd ∈ Q and bc + ad ∈ Q. Therefore S is closed under + and ·.
Next we need to show that S is an abelian group under +. Since R under + is abelian, it
suffices to show that S is a subgroup of R under +. (i)√We have already shown that S is
closed under +. (ii) Certainly 0 ∈ S √
because 0 = 0 + 0 2. (iii)
is closed under
√ Finally, S √
additive inverses, because given a + b 2 ∈ S, we have −(a + b 2) = −a − b 2, which is in
S because −a, −b ∈ Q.
The remaining ring axioms are all inherited from R. Multiplication of real numbers is associative, and the distributive laws hold in R. Thus S is a ring.
(b) To show that S is a field, we first observe that S is commutative since multiplication of
real numbers is commutative. Then we need to show that S ∗ = S \ {0} is a group under
multiplication. (i) It is certainly closed under ·, since the product of non-zero real numbers
is
√
always non-zero (and S is already closed under · by (a)). (ii) 1 ∈ S ∗ , since 1 = 1 + 0 2 ∈ S.
√
(iii) To see that S ∗ is closed under multiplicative inverses, we let a + b 2 ∈ S ∗ , with a, b ∈ Q.
Then
√
1
1
a−b 2
√ =
√ ·
√
a+b 2
a+b 2 a−b 2
√
a−b 2
= 2
a − 2b2
√
a
b
= 2
2.
−
a − 2b2 a2 − 2b2
b
a
We see that this is an element of S because a2 −2b
2 ∈ Q and a2 −2b2 ∈ Q.
√
√
√
(One subtle point is that a + b 2 6= 0 √
implies that a − b 2 6= 0, since 2 is irrational, so it
was ok to multiply and divide by a − b 2 in the calculation above.)