Download Math 236H Final exam

Math 236H
Final exam
May 6, 2008
Be sure to write your name on your bluebook. Use a separate page (or pages) for each problem.
Show all of your work.
1. (15 points) Prove that the symmetric group S4 is generated by the three transpositions (12),
(23) and (34).
Solution: We know that S4 is generated by 6 transpositions: the three listed above along
with (13), (14) and (24). Thus we need to show that each of these three is a product of
the three listed above. Let the four letters being permuted by A, B, C and D. Then we
have
(A, B, C, D)
(A, B, C, D)
(A, B, C, D)
(12)
(B, A, C, D)
(23)
(A, C, B, D)
(23)
(B, C, A, D)
(12)
(C, B, A, D)
(12)
(B, A, C, D)
(34)
(A, C, D, B)
(23)
(B, C, A, D)
(34)
(A, D, C, B)
(34)
(B, C, D, A)
(23)
(B, D, C, A)
(12)
(D, B, C, A)
These three columns show that (13), (24) and (14) are each products of the three generators.
2. (5 points) Eisenstein’s criterion says that the polynomial
a(x) = xn + a1 xn−1 + a2 xn−2 . . . + an ∈ Z[x]
is irreducible (over Z) if
(i) each coefficient ai is divisible by a prime p and
(ii) an is not divisible by p2 .
Give an example of a polynomial satisfying (i) but not (ii) that is reducible.
Solution: One example is (x + p)2 = x2 + 2px + p2 .
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Final exam
3. (15 points) Let p be an odd prime, and let
f (x) =
x(x(p−1)/2 + 1)(x(p−1)/2 + p − 1)
.
2p
Prove that f (x) is an integer whenever x is.
Solution: This amounts to showing that the numerator is always even and always divisible by p. Its two factors besides x differ by p − 2, which is odd, so one of them must be
even, so their product is divisible by 2. For divisbility by p, note that the numerator is
x(x(p−1)/2 + 1)(x(p−1)/2 + p − 1) = xp + px(p+1)/2) + (p − 1)x ≡ xp − x
mod p,
and xp − x is divisible by p by Fermat’s Little Theorem.
4. (10 points) Find a symmetric group Sn with an element of order greater than 2n and identify
the element.
Solution: S9 has en element of order 20, namely (12345)(6789). There are many other
examples.
5. (10 points) Let G be a group of order pq where p and q are disinct primes. Show that every
proper subgroup of G is cyclic.
Solution: If H is a proper subgroup of G, its order must be less than pq and divide pq,
so it must be 1, p or q. Hernce H is either trivial or of prime order, so it is cyclic.
6. (15 points) List the even permutations of order 2 in S4 and say how many there are in S5
and S6 .
Solution: The only even permutations of order 2 in S4 are the three double transpositions, (12)(34), (13)(24) and (14)(23). Double transpositions are also the only such
permutations in S5 and S6 . (There are triple transpositions in S6 which have order 2, but
they are odd.) In S5 there are three for each of the five subsets with 4 elements, making
15 in all. In S6 there are three for each of the fifteen subsets with 4 elements, making 45
in all.
7. (10 points) You are making necklaces with 7 beads each, and each bead can be any one
of n colors. Use Burnside’s formula to find the number r(n) of distinct necklaces. Find the
smallest n such that this number exceeds 1000.
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Math 236H
Final exam
Solution: The relevant group here is the dihedral group D14 and the set X of all possible
P
necklaces has n7 elements. The number of orbits r(n) is g∈D14 |Xg |/14. The find r, note
that
• The identity element fixes all n7 necklaces.
• Each of the six nontrivial rotations fixes the n necklaces in which all beads have the
same color.
• Each of the seven reflections fixes one bead and interchanges three pairs of beads,
so it fixes n4 necklaces.
Hence the number of orbits is
r(n) =
n7 + 7n4 + 6n
n(n3 + 1)(n3 + 6)
=
14
14
so r(2) = 2 · 9 · 14/14 = 18, r(3) = 3 · 28 · 33/14 = 198 and r(4) = 4 · 65 · 70/14 = 1300.
Note that r(n) = f (n), where f is the function of Problem 3 for p = 7. f (n) is the number
of necklaces for an odd prime number p of beads and n colors.
8. (10 points) Describe the commutator subgroup C(G) of every simple
a. abelian group
b. nonabelian group
Solution:
a. In an abelian group, all commuators are trival, so C(G) is trivial
b. The commutator subgroup is normal, so if G is simple, C(G) is either G or {e}. It
cannot be the latter since the G is nonabelain, so C(G) = G
9. (10 points) Prove that the intersection of two normal subgroups of G is a normal subgroup.
Solution: Let H1 and H2 be normal subgroups of G. Let g ∈ G and h ∈ H1 ∩ H2 . Then
ghg −1 ∈ H1 since h ∈ H1 , and similarly ghg −1 ∈ H2 . Hence ghg −1 ∈ H1 ∩ H2 , so H1 ∩ H2
is normal.
10. (10 points) Determine the number of elements of order 4 in
a. C8 × C8
b. C4 × C4 × C4
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Solution:
a. All elements of order 4 or less are in the subgroup C4 × C4 , which has order 16, and
all elements of order 2 or less are in the subgroup C2 × C2 , which has order 4, so
the number of lements of order 4 is 16 − 4 = 12.
b. An element of C4 × C4 × C4 has order 4 if it is not contained in C2 × C2 × C2 . The
number of such elements is 43 − 23 = 56.
11. (10 points) Let n be an integer such that 6n − 1 and 6n + 1 are both primes. (Such primes
are called twin primes.) Show that every group G of order 36n2 − 1 is cyclic.
Solution: Let p = 6n − 1 and q = 6n + 1, so |G| = pq. We will use the third Sylow
theorem to show that there is exactly one subgroup of order p and one of order q. The
number of p-Sylow subgroups divides pq (so it is 1, p, q or pq) and is congruent to 1
modulo p. The only number that fits this description is 1. Hence the p-Sylow subgroup
is normal, so there is a homomorphism φ1 from G onto the quotient group, which must
be Cq . Similarly, there is only one q-Sylow subgroups and hence a homomorphism φ2 to
Cp . Using these we get a homomorphism to Cp × Cq = Cpq which is onto and therefore
an isomorphism.
12. (15 points) Prove that every finite integral domain is a field.
Solution: See Theorem 19.11 on page 181 of Fraleigh.
13. (10 points) Describe a nonabelian group of order 55. Hint: Consider the group of 2 × 2
matrices (under multiplication) of integers modulo 11 of the form
"
M=
a b
0 1
#
with a 6= 0.
Solution: The set of matrices of this form is a group under multiplication since
"
a b
0 1
#"
and
"
a b
0 1
a0 b0
0 1
#
#−1
"
=
"
=
aa0 ab0 + b
0
1
a−1 −a−1 b
0
1
#
#
There are 10 possible values of b and 11 possible values of b, so there are 110 matrices of
this form. We to find a 5-element subset of a-values that is closed under multiplication,
so we need an element whose 5th power is 1. One such element is 3, so the subset is
{3, 9, 5, 4, 1}. Hence our subgroup consists of matrices M in which a is a power of 3.
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Math 236H
Final exam
14. (10 points) Prove that if n is an odd integer, then n2 ≡ 1 modulo 8 and n4 ≡ 1 modulo 16.
Solution: Let n = 2k + 1. Then
n2 = (2k + 1)2 = 4k 2 + 4k + 1
= 4k(k + 1) + 1
and k(k + 1) is even since it is the product of two consecutive integers. This means that
n2 is congruent to 1 modulo 8. Since n2 = 8s + 1, we have
We also have
n4 = (8s + 1)2 = 64s2 + 16s + 1
≡ 1
mod 16.
15. (15 points) For which integers r is f (x) = x4 + rx3 + x + 3 reducible over the integers? Prove
your answer.
Solution: If f (x) is reducible, it is either the product of two quardatic factors, or of a
linear factor and a cubic one. In the former case the product of the constant terms must
by 3, so we have
f (x) = (x2 + ax + 1)(x2 + bx + 3)
or
f (x) = (x2 + ax − 1)(x2 + bx − 3).
In the first of these, we have
f (x) = (x2 + ax + 1)(x2 + bx + 3) = x4 + (a + b)x3 + (4 + ab)x2 + (3a + b)x + 3
so ab = −4 and 3a + b = 1. The only possibility here is (a, b) = (−1, 4), so r = a + b = 3
and
x4 + 3x3 + x + 3 = (x3 + 1)(x + 3) = (x + 1)(x2 − x + 1)(x + 3).
In the second case of quadratic factors we have
f (x) = (x2 + ax − 1)(x2 + bx − 3) = x4 + (a + b)x3 + (−4 + ab)x2 − (3a + b)x + 3
so ab = 4 and 3a + b = −1. These lead to 3a2 + a + 4, which has no integral solution.
Now suppose f (x) has a root. It must be ±1 or ±3. We have
f (1) = 1 + r + 1 + 3 = 5 + r
f (−1) = 1 − r − 1 + 3 = 3 − r
f (3) = 81 + 27r + 3 + 3 = 87 + 27r
f (−3) = 81 − 27r − 3 + 3 = 81 − 27r
This gives us one new value of r, namely r = −5, and
x4 − 5x3 + x + 3 = (x − 1)(x3 − 4x2 − 4x − 3)
Hence f (x) is reducible only for r = 3 and r = −5.
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