Download Math 236H Final exam

Math 236H
Final exam
May 8, 2012
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 if n is an odd integer not divisible by 3, then n2 ≡ 1 modulo 24.
Solution: Let n = 6k ± 1. Then
n2 = (6k ± 1)2 = 36k 2 ± 12k + 1
= 12k(3k + 1) + 1
and k(3k + 1) is even since both factors cannot be odd. This means that n2 is congruent
to 1 modulo 24.
2. (10 points) Determine the number of elements of order 25 in
a. C125 × C125
b. C25 × C25 × C25
Solution:
a. All elements of order 25 or less are in the subgroup C25 × C25 , which has order 625,
and all elements of order 5 or less are in the subgroup C5 × C5 , which has order 25,
so the number of lements of order 25 is 625 − 25 = 600.
b. An element of C25 × C25 × C25 has order 4 if it is not contained in C5 × C5 × C5 .
The number of such elements is 253 − 53 = 15625 − 125 = 15500.
3. (15 points) Prove that every finite integral domain D is a field.
Solution: Suppose |D| = n and
D = {0, 1, a1 , a2 , . . . an−2 } ,
so the set of nonzero elements in D is
D∗ = {1, a1 , a2 , . . . an−2 }
For x ∈ D∗ , consider the set
xD∗ = {x, xa1 , xa2 , . . . xan−2 } .
Since D is a domain, this set does not contain 0. Its elements are all distinct since
xai = xaj for i 6= j would mean x(ai − aj ) = 0, which would make x a zero divisor. Hence
xD∗ has all n − 1 nonzero elements of D, including 1. This means that x has an inverse,
so D is a field.
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Final exam
4. (10 points) Let f (x) = x4 + 5x2 + rx + 1 ∈ Z[x]. For which integers r is f (x) irreducible?
Prove your answer.
Solution: If a is a zero of f (x), then x − a is a factor of f (x), f (a) = 0 and a must be
±1 since the constant term in f (x) is one. Since f (1) = 7 + r, and f (−1) = 7 − r, f (x)
has a zero if r = ±7. The factorizations are
x4 + 5x2 + 7x + 1 = (x + 1)(x3 − x2 + 6x + 1)
and
x4 + 5x2 − 7x + 1 = (x − 1)(x3 + x2 + 6x − 1).
Now suppose that f (x) has two quadratic factors,
f (x) = (x2 + ax + b)(x2 + cx + d)
= x4 + (a + c)x3 + (b + d + ac)x2 + (ad + bc)x + bd.
The coefficient of x3 is zero, so c = −a and we have
f (x) = x4 + (b + d − a2 )x2 + a(b − d)x + bd.
Since bd = 1, b = d = ±1, and the coefficient of x is zero. This means that r = 0 and the
coefficient of x2 is −a2 ± 2 = 5, or a2 = −5 ± 2 < 0, but there is no integer a that satisfies
this condition.
Therefore, f (x) is irreducible unless r = ±7.
5. (10 points) Prove that in a finite G-set X, if g and g 0 are conjugate elements in G, then
their fixed point sets Xg and Xg0 have the same cardinality.
Solution: Since g 0 is conjugate to g, g 0 = hgh−1 for some h ∈ G. Now suppose x ∈ Xg .
Then
g 0 (h(x)) = hgh−1 h(x) = hg(x) = h(x),
so h(x) ∈ Xg0 . Similarly, if x0 ∈ Xg0 , then h−1 (x0 ) ∈ Xg . Hence the action of h gives a
one-to-one correspondance between Xg and Xg0 , so |Xg | = |Xg0 |.
6. (20 points) You are painting octahedral blocks and you have n colors to choose from for
each of the 8 faces. How many distinguishable blocks can be made in this way? How big does
n have to be in order to get at least 50 distinguishable blocks?
Solution: We will use Burnside’s formula
X
r=
|Xg |/|G|
g∈G
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Math 236H
Final exam
to find the number of orbits r. The group G here is the symmetry group of the octahedron,
which is isomorphic to S4 , and the set X has n8 elements. Conjugate elements in the
group will have fixed points sets of the same size, and there are five such classes:
• The identity element, which fixes all n8 colorings.
• The six 4-cycles, each of which corresponds to a π/2 (90 degrees) rotation about a
vertex. The four faces toching the given vertex must all have the same color, and
the same goes for the four faces touching the opposite vertex. This means there are
n2 colorings that are fixed.
• The three double transpositions, which are squares of 4-cycles. Each corresponds to
π (180 degrees) rotation about a vertex, and each swaps four pairs of faces. There
are n4 colorings that are fixed.
• The eight 3-cycles. Each corresponds to a 2π/3 (120 degrees) rotation about the
center of a face. That face and its opposite are fixed. The remaining six faces get
permuted by a pair of 3-cycles. Thus there are 4 orbits among the 8 faces, so n4
colorings are fixed.
• The six transpositions. Each corresponds to a π (180 degrees) rotation about the
center of an edge. Four pairs of faces get swapped, so n4 colorings are fixed.
It follows that Burnside’s sum is
r =
=
n8 + 6n2 + 3n4 + 8n4 + 6n4
24
n8 + 17n4 + 6n2
24
This number is 23 for n = 2 and 333 for n = 3.
7. (15 points) Recall that the third Sylow theorem says that if |G| = ps where p is prime and
does not divide s, then the number kp of subgroups of order p divides s and is congruent to
1 modulo p. Use it to prove that every group G of order 665 = 5 · 7 · 19 is cyclic.
Solution: For each of the three primes p dividing 665, there is a unique subgroup Cp
of order p, so it is normal. The order of quotient group is the product of the other two
primes, and in each case Theorem 37.7 applies and says that it is abelian. (It is not true
in general that a group H whose order is the product of 2 primes p and q is abelian.
For example, |S3 | = 2 · 3 but S3 is not abelian. However such an H is abelian if neither
prime is congruent to 1 modulo the other one.) This means that each Cp contains the
commutator subgroup C(G). Hence C(G) is in the intersection of the three subgroups of
prime order, so it is trivial. This means that G is abelian and hence cyclic.
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Final exam
An alternate approach is to show there is an element of order 665, which would make the
group cyclic. This can be done by showing that not all elements of G have order less than
665. We know there is a single element of order 1 and p − 1 elements of order p for each of
the three primes p. There can be at most one subgroup of order 35 = 5 · 7, 95 = 5 · 19 or
133 = 7 · 19, because in each case multiple subgroups of any of these orders would lead to
multiple subgroups of prime order. Thus the number of elements of order pq is less that
pq. The order of any element must divide 665, and there are fewer than 665 elements of
order less than 665.
8. (10 points) Find the largest integer m which divides n17 − n for all integers n and prove
your answer.
Solution: We will use Fermat’s little theorem, which says that if a prime p does not
divide n, then np−1 ≡ 1 modulo p. Since
n17 − n = n(n16 − 1),
if p − 1 divides 16, then n17 − n is divisible by p. The primes p for which this is true are
2, 3, 5 and 17. This means that n17 − n is always divisible by their product, 510. For any
other prime q there is an n such that neither n nor n16 − 1 are divisible by q. For any
prime p, the number p17 − p = p(p16 − 1) is divisible by p, but not by p2 . This means m
is not divisible by the square of any prime (it is square free), so our m is 510.
9. (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.
10. (10 points) Let G be a finite group with a subgroup H such that |G| = 2|H|.
(a) Prove that if a ∈ G is not in H, then a2 ∈ H. (Hint: it suffices to show that a2 H 6= aH.)
(b) Prove that if a is not in H, then the order a is even. (Hint: Show that an is not in H
for any odd integer n.)
Solution:
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Math 236H
Final exam
(a) G has two left H-cosets, namely H itself and its compliment G − H. If aH = a2 H,
then H = aH, but if a is not in H, then aH = G − H. Hence aH 6= a2 H, so
a2 H = H, which means that a2 ∈ H.
(b) For an odd integer n = 2m + 1, an = a · a2m ∈ aH, so an is not in H. This means
the order of a cannot be odd.
11. (10 points) Find all primes p such that x + 3 is a factor of f (x) = x6 + x3 + 1 in Z/p[x].
Solution: x + 3 is a factor of f (x) in Z/p[x] iff f (−3) ≡ 0 modulo p. We have
f (−3) = 36 − 33 + 1 = 729 − 27 + 1 = 703 = 19 · 37
so the primes are 19 and 37.
12. (15 points) Let p = 2s + 1 be an odd prime bigger than 3 (so s > 1), and let
f (x) =
x(xs + 3p − 1)(xs + 1)
.
3p
Prove that f (x) is an integer whenever x is.
Solution: This amounts to showing that the numerator is always divisible by three and
always divisible by p. Modulo three it is
x(xs − 1)(xs + 1)
If x is not divisible by 3, then xs ≡ ±1 modulo 3, so the product is always divisible by 3.
For divisibility by p, note that the numerator is
x(xs + 3p − 1)(xs + 1) ≡ x(xs − 1)(xs + 1) ≡ xp − x
mod p,
and xp − x is divisible by p by Fermat’s Little Theorem.
13. (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.
14. (10 points) Prove that the intersection of two normal subgroups of G is a normal subgroup.
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Final exam
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.
15. (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.
16. (2 points) What famous historical event happened on this date between 60 and 70 years
ago?
Solution: VE Day, the formal end of World War II in Europe, was May 8, 1945.
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