Download Final Exam Review Problems and Solutions

Math 280 Final Exam Practice
Final Exam: Friday, December 11, 8:00-10:00 am
1. Decide which of the following are equivalence relations on the set S. If you believe they are, prove it
by showing the properties are all satisfied. If you believe they are not, prove it by showing a specific
example where one of the properties fails.
(a) S is the set of integers, a ∼ b if a < b
(b) S is the set of integers, a ∼ b if a ≡ b mod m for any natural number m
2. Decide which of the following are groups. If you believe they are, prove it by showing the properties
are all satisfied. If you believe they are not, prove it by showing a specific example where one of the
properties fails.
(a) G is the set of integers under subtraction.
(b) G is the set of 3 × 3 matrices with determinant 1 or −1 under matrix multiplication.
3. Make a Cayley table for the following groups:
(a) The group Z4 under addition mod 4
(b) The group U (8) = {1, 3, 5, 7} under multiplication mod 8
4. Prove that the following sets H are subgroups of the group G:
(a) H is the set of even numbers, G is the group of integers under addition
(b) H is the set of numbers of the form
numbers under multiplication
1
2m
for integers m, G is the group of nonzero rational
5. Show that if g is an element of a group G, then hgi is a subgroup of G.
6. Let a, b, c be elements of a group with |a| = 6 and |b| = 7. Express (a4 c−2 b4 )−1 without any negative
exponents.
7. Prove that, for group elements a and b, (ab)2 = a2 b2 if and only if ab = ba.
8. Decide whether each of the following statements is true or false. If the statement is true, prove it.
Otherwise, show it is false by providing a counterexample.
(a) Let ∼ be an equivalence relation on a set S. If for elements a, b, c ∈ S we have c ∼ a and c ∼ b,
then a ∼ b.
(b) If b ∈ [a], then [b] = [a].
(c) There are no elements with finite order in an infinite group.
(d) The only generator of the group {1, −1, i, −i} under multiplication is 1.
T
(e) If H and K are subgroups of a group G, then H K is also a subgroup of G.
(f) For a group element a, it is possible that |a| =
6 |a−1 |.
9. Explain why G = S4 and H = S6 are not isomorphic.
10. Write each of the following permutations as a product of disjoint cycles:
(a) (1 3 4 2)(1 7 6 2)(2 5 9 8)
(b) (1 5 3 4 2)(1 5 2 4)
(c) (1 3)(2 5)(3 1)(2 5)(4)
1
11. Find the order and inverse of each permutation in the previous problem.
12. Find two groups of order 6 that are not isomorphic, and explain how you know they are not isomorphic.
13. Find the external direct product of Z2 ⊕Z5 (list the elements explicitly). What group is this isomorphic
to?
14. Prove that if H is a subgroup of G and h ∈ H, then hH = H = Hh.
15. Show that if G and H are isomorphic, then |G| = |H|.
16. Explain why Z17 is cyclic (you don’t need to prove this, just state a theorem that tells you it is true).
How many generators does it have?
17. Let G = {1, i, −1, −i} under complex multiplication (recall that G is a group!). Consider the map
φ : G → Z4 defined by
φ(1) = 0,
φ(i) = 1,
φ(−1) = 2,
φ(−i) = 3
Show that φ is an isomorphism.
18. Prove that if G and H are isomorphic groups, and G is Abelian, then H must be Abelian.
19. For a group G, define a map φ : G → G by φ(g) = g −1 . Is φ always an isomorphism? What property
of G would guarantee that φ is an isomorphism?
20. Let G1 , G2 , G3 be groups. Suppose φ : G1 → G2 is an isomorphism, and ψ : G2 → G3 is an
isomorphism. Show that ψ ◦ φ : G1 → G3 is an isomorphism. (This proves the transitive property
holds for isomorphism: if G1 ≈ G2 and G2 ≈ G3 , then G1 ≈ G3 , as we might expect).
21. Decide whether or not Z12 is an internal direct product of its subgroups H and K when
(a) H = h3i = {0, 3, 6, 9} and K = h2i = {0, 2, 4, 6, 8, 10}
(b) H = h3i = {0, 3, 6, 9} and K = h4i = {0, 4, 8}
22. Find all possible Abelian groups of order 180. Which is isomorphic to Z180 ?
23. Decide whether or not each of the following maps is a homomorphism, an isomorphism, or neither.
(a) φ : G → G, where G is any group and
φ(x) = x
(b) φ : G → G, where G is any group with identity e and
φ(x) = e
(c) φ : Z → Z2 defined by
(
0 if x is even
φ(x) =
1 otherwise
24. Suppose G and H are groups, and f : G → H is a homomorphism. Show that f preserves the identity:
f (eG ) = eH .
25. Let R be a ring with ideals I and J. Show that I ∩ J is an ideal of R.
2
26. Let m and n be arbitrary integers.
(a) Describe the principal ideals hmi and hni.
(b) Calculate hmi ∩ hni.
(c) Calculate hmi + hni.
27. Let R be a (nontrivial) commutative ring with unity. Show that R is an integral domain if and only if
whenever ab = ac for a, b, c ∈ R and a 6= 0, we have b = c. That is, R is an integral domain if and
only if the cancellation law holds in R.
28. Decide which of the following are ring homomorphisms:
(a) φ : Z3 → Z12 is defined by φ(x) = 4x.
(b) φ : R × R → R is defined by φ(x, y) = xy.
(c) φ : R[x] → R is defined by φ(f (x)) = f (1). (Recall R[x] is the ring of polynomials with real
coefficients.)
29. Give an example of a set that:
(a) is commutative ring with unity, but not an integral domain.
(b) is a commutative ring, but does not contain a unity element.
(c) is an integral domain, but not a field.
(d) is an infinite field.
(e) is a finite field.
(f) is not commutative, has zero divisors, and does not contain multiplicative inverses, but is a ring
with unity.
30. Find polynomials equivalent to those listed below in the indicated field. Then, find any zeros of the
polynomial in that field.
(a) 5x2 − 13x + 9, in Z3
(b) 7x3 + 14x + 1, in Z7
31. Find an example of a ring homomorphism from R[x] to itself. That is, a map φ such that
φ(f (x) + g(x)) = φ(f (x)) + φ(g(x))
and
φ(f (x)g(x)) = φ(f (x))φ(g(x))
for any polynomials f (x), g(x) with real coefficients.
3
Math 280 Final Exam Practice Solutions
1. (a) This is not an equivalence relation, since symmetry does not hold. For instance, 3 < 4, so 3 ∼ 4
by this relation, but 4 is not less than 3, so 4 is not related to 3.
(b) Symmetric: a ≡ a mod m for any a, since a = a + 0m.
Reflexive: Suppose a ≡ b mod m, so we can write a = b + km for some integer k. Then
a − b = km, so b − a = −(a − b) = −(km) = (−k)m. Then b = a + (−k)m, so we have b ≡ a
mod m.
Transitive: Suppose a ≡ b mod m and b ≡ c mod m. Then there are integers k and l so that
a = b + km and b = c + lm. Then a = (c + lm) + km = c + (l + k)m, and so a ≡ c mod m.
Then if a ∼ b and b ∼ c, we have a ∼ c.
This is an equivalence relation (the fact that we refer to numbers being equivalent mod m
certainly implies there is an equivalence relation!).
2. (a) This is not a group, and is one of the cases where associativity fails. Note that
1 − (2 − 3) = 1 − (−1) = 2
but
(1 − 2) − 3 = (−1) − 3 = −4
so associativity does not hold.
We do, however, have an identity (0) and everything has an inverse.
(b) Associativity: Inherited from associativity of matrix multiplication.
Identity: The identity matrix is our multiplicative identity (there was a reason we call it the
identity matrix!)
Inverses: If a matrix has determinant 1 or -1, then it must be invertible. The only requirement
is that the determinant is nonzero.
So this is a group (it’s the orthogonal group).
Note that we can’t form a group from the set of matrices with just any determinant, like 2, since
the identity isn’t included. Even if we take the matrices with determinant 1 and 2, we wouldn’t
have a group, since we wouldn’t have closure. If det(A) = 2 and det(B) = 2, then det(AB) = 4,
which means AB isn’t in the set.
3. (a)
+4
0
1
2
3
0
0
0
2
3
1
0
1
3
0
2
0
2
0
1
3
0
3
1
2
(b)
×8
1
3
5
7
1
1
3
5
7
3
3
1
7
5
5
5
7
1
3
7
7
5
3
1
4. (a) Let a, b ∈ H, so a and b are even numbers. The inverse of a under addition would be −a, which
is also even, and is therefore in H. Plus, the product of two even numbers is even, so ab is also
in H. By the two-step subgroup test, H must be a subgroup.
1
(b) Let a, b ∈ H, so a = 21m and b = 21n for some integers n, m. The inverse of a under multiplication
would be 2m , which can also be written
a−1 = 2m = 2−(−m) =
1
2−m
and since −m is an integer, then a−1 is in H. Finally, note that
1
1
1
ab =
= m+n
m
n
2
2
2
and since m + n is an integer, then ab is in H. By the two-step subgroup test, H must be a
subgroup.
5. We’ll stick with the multiplicative notation here, so hgi is the set of all elements of the form g n for
integers n. Certainly 0 is an integer, so e = g 0 ∈ hgi, and we have an identity element. If g m ∈ hgi,
then so is g −m , so everything has an inverse. Since G is a group, it has the associative property, which
hgi inherits. So hgi is a group. Obviously it’s contained in G and under the same operation, so it’s a
subgroup of G.
We could also use the subgroup test here. If a, b ∈ hgi, then a = g n and b = g m for some integers
n, m. Then
ab−1 = g n g −m = g n−m
is also in hgi, so it’s a subgroup.
6. We can write
(a4 c−2 b4 )−1 = (b4 )−1 (c−2 )−1 (a4 )−1
(we can’t assume the group is Abelian, so use the “shoes and socks” rule). Now,
(a4 c−2 b4 )−1 = (b4 )−1 (c−2 )−1 (a4 )−1 = b−4 c2 a−4
To get rid of the negative exponents, we’ll use the definition of the order of an element. We can say
a has order 6 if a6 = e. If we multiply both sides of the equation by a−1 , we get a5 = a−1 . If we keep
going, a4 = a−2 , a3 = a−3 , a2 = a−4 , and a = a−5 . In general, if x has order m, then x−k = xm−k .
So if b has order 7, then b−4 = b7−4 = b3 , and we have
(a4 c−2 b4 )−1 = (b4 )−1 (c−2 )−1 (a4 )−1 = b−4 c2 a−4 = b3 c2 a2
7. (⇒) Assume (ab)2 = a2 b2 . Expanding the left side and using the fact that a group has the associative
property,
(ab)2 = (ab)(ab) = abab = a2 b2
Now we’ll multiply by a−1 and b−1 (which must be in the group!) on the left and right sides, as
needed, to get ab = ba:
bab = ab2
multiply on left by a−1
ba = ab
multiply on right by b−1
(⇐) Assume ab = ba for all elements a, b in the group. This means that the group is Abelian, so
(ab)2 = (ab)(ab) = abab = aabb = a2 b2
8. (a) True. Suppose a, b, c ∈ S, and ∼ is an equivalence relation on S such that if c ∼ a and c ∼ b,
then a ∼ b. Then if c ∼ a, we must have a ∼ c by the symmetric property. If we also have c ∼ b,
then by the transitive property,
a ∼ c and c ∼ b
2
⇒
a∼b
(b) True. Equivalence classes are disjoint, so if there’s an element in common, they have to be
identical.
(c) False. The identity element always has order 1, which is finite, even if the group is infinite. There
may not be others, but there’s at least that one element.
(d) False. In fact, 1 isn’t a generator of the set, since there’s no power of 1 that we can take to get
i, −i, or −1. But, this is a cyclic group, since
i1 = i,
i2 = −1,
i3 = −i,
i4 = 1
So there is a generator, it just isn’t 1.
T
(e) True. We’ll use the subgroup test. Suppose a, b ∈ H K. Then a and b must be in both H
and K, by definition of intersection. Since H and KTare groups, they have the inverse property,
so a−1 must be in both H and K, and T
hence in H K. Finally, ab must be in both H
T and K
(since they’re groups!) and so ab ∈ H K. Then by the two-step subgroup test, H K is a
subgroup of G.
This can be extended: any arbitrary intersection of subgroups of a group G is another subgroup.
Do you think this is true for unions of subgroups?
(f) False. This is not possible, which we can prove by showing that |a| = |a−1 |. We’ll do this by
showing that |a| ≤ |a−1 |, and that |a−1 | ≤ |a|:
Suppose |a| = n, so by definition, an = e, and n is the smallest positive integer for which this is
true. Let |a−1 | = k. Now
(a−1 )n = (an )−1 = e−1 = e
So if we raise a−1 to the power of n, we get the identity. It feels like we’re done, but remember
that the order of a−1 must be the smallest positive value k for which (a−1 )k = e. All we’ve really
done is show that k divides n. For instance, if a has order 6, then we can now say that (a−1 )6 = e,
but this doesn’t rule out the possibility that (a−1 )2 = e, since a6 = ((a−1 )2 )3 = e3 = e. So we
only know that k ≤ n.
Note that
ak = ((a−1 )−1 )k = ((a−1 )k )−1 = e−1 = e
so if we raise a to the power of k, we get the identity. This means n must divide k, so n ≤ k.
If n ≤ k and k ≤ n, then the only possibility is that n = k. So the order of an element is equal
to the order of its inverse.
9. Since G has order 4! = 24 and H has order 6! = 720, then the two groups cannot be isomorphic, as
isomorphic groups always have the same order.
10. (a) (1 7 6)(2 5 9 8 3 4)
(b) (1 3 4 5)(2)
(c) (1)(2)(3)(4)(5)
11. |(1 7 6)(2 5 9 8 3 4)| = 6,
|(1 3 4 5)(2)| = 4,
|(1)(2)(3)(4)(5)| = 1,
((1 7 6)(2 5 9 8 3 4))−1 = (1 6 7)(2 4 3 8 9 5)
((1 3 4 5)(2))−1 = (1 5 4 3)(2)
((1)(2)(3)(4)(5))−1 = (1)(2)(3)(4)(5)
12. We only have a few examples of groups with order 6. S3 , the permutations on 3 numbers, has 6
elements, as does Z6 . But Z6 is Abelian and S3 is not, so we know they are not isomorphic.
13. Z2 ⊕ Z5 = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4)}
This group is isomorphic to Z10 , since 2 and 5 are relatively prime and 2 · 5 = 10.
3
14. There are several ways to answer this - see if you can find another one!
Since H is a subgroup, it must contain the identity, e. The left coset of H containing e is clearly
equal to H, as is the right coset:
eH = {eg | g ∈ H} = {g | g ∈ H} = H = {ge | g ∈ H} = He
Now suppose h ∈ H with h 6= e. Again, since H contains the identity, the left coset of H containing
h must contain the product he = h, so h really is in the left coset hH. Since left cosets are equal or
disjoint, and h must also be in eH, then eH = hH = H. We can say the same thing about the right
coset, Hh.
15. Suppose φ is an isomorphism from G to H. Then φ is one-to-one, which means two different elements
of G cannot map to the same element in H. Then the order of G can’t be larger than the order of
H. On the other hand, φ is onto, so there has to be some element of G that maps to each element
in H; this implies that the order of H can’t be larger than the order of G. Therefore, the order of G
and H must be the same.
16. Every group of prime order is cyclic, which would include Z17 . Since the order of an element divides
the order of the group, then every element in Z17 must have order 1 or 17. Only the identity has order
1, so the remaining 16 elements each have order 17. If the order of an element equals the order of the
group, it is a generator. So, Z17 is cyclic, and has 16 generators.
17. The map is certainly one-to-one and onto; we just need to make sure that it preserves the properties
of the group. If you check all the possibilities, such as
φ(1 · i) = φ(i) = 1 = 0 + 1 = φ(1) + φ(i)
you’ll see the map truly is an isomorphism.
18. Suppose G is Abelian. We want to show the same is true for H, so if h, h0 ∈ H, then hh0 = h0 h.
Since G and H are isomorphic, there is some isomorphism φ : G → H. By definition, this map φ is
onto, so for any h ∈ H, there is some g ∈ G so that φ(g) = h. Similarly, there is some g 0 ∈ G so that
φ(g 0 ) = h0 . Now
hh0 = φ(g)φ(g 0 ) = φ(gg 0 ) = φ(g 0 g) = φ(g 0 )φ(g) = h0 h
19. Since the inverse of a group elements is unique, and every element in a group has one, then the map
is one-to-one and onto. We just need to see if it preserves the group operation; that is, does
φ(g1 g2 ) = φ(g1 )φ(g2 )
Let g1 , g2 ∈ G. Then
φ(g1 g2 ) = (g2 g1 )−1 = g2−1 g1−1
while
φ(g1 )φ(g2 ) = g1−1 g2−1
Since these two products are not necessarily equal, φ may not be an isomorphism. It would be if and
only if G is Abelian, in which case
g1−1 g2−1 = g2−1 g1−1
4
20. We can assume φ and ψ are one-to-one, onto, and preserve the operations of the groups. We just
need to make sure the same holds for their composition, ψ ◦ φ. Suppose g, g 0 ∈ G1 . Then
ψ(φ(g)) = ψ ◦ φ(g) = ψ ◦ φ(g 0 ) = ψ(φ(g 0 ))
implies φ(g) = φ(g 0 ), since ψ is one-to-one, which in turn implies that g = g 0 , since φ is one-to-one.
Next, suppose y ∈ G3 . We want to show there is some x ∈ G1 such that ψ ◦ φ(x) = ψ(φ(x)) = y.
Since φ and ψ are both isomorphisms, they are invertible. Then we can let
x = φ−1 (ψ −1 (y))
Finally, we’ll show the composition preserves the group operations. Note that
ψ ◦ φ(gg 0 ) = ψ(φ(gg 0 )) = ψ(φ(g)φ(g 0 ))
since φ is an isomorphism. Similarly, ψ is an isomorphism, so
ψ ◦ φ(gg 0 ) = ψ(φ(g)φ(g 0 )) = ψ(φ(g))ψ(φ(g 0 ))
as required.
21. (a) The intersection of H and K contains more than the identity, so this would not be an internal
direct product.
(b) We need to check the three requirements for internal direct products:
i. H and K intersect only at the identity, 0.
ii. Every element in Z12 can be written as the sum, mod 12, of elements in H and K. For
instance, 1 = 9 + 4 and 2 = 6 + 8.
iii. Both H and K are normal in Z12 . We know this because Z12 is Abelian, so all of its
subgroups are normal.
22. Start by finding the prime factorization: 180 = 22 32 5. The 4 possible Abelian groups of order 180 are
Z2 ⊕ Z2 ⊕ Z3 ⊕ Z3 ⊕ Z5
Z4 ⊕ Z3 ⊕ Z3 ⊕ Z5
Z2 ⊕ Z2 ⊕ Z9 ⊕ Z5
Z4 ⊕ Z9 ⊕ Z5
The last one is isomorphic to Z180 , since 4, 9, and 5 are pairwise relatively prime.
23. (a) It is certainly a homomorphism, since
φ(xy) = xy = φ(x)φ(y)
for all x, y ∈ G. It is also an isomorphism, since it is one-to-one and onto.
(b) This is a homomorphism:
φ(xy) = e = ee = φ(x)φ(y)
for all x, y ∈ G. It is not an isomorphism, since it is neither one-to-one nor onto (unless G is the
trivial group, which we cannot assume here).
5
(c) Let x, y ∈ Z. We need to check that φ preserves the group operations; that is,
φ(x + y) = φ(x) +2 φ(y)
Now, if x and y are both even, so is x + y, so
φ(x + y) = 0 = 0 +2 0 = φ(x) +2 φ(y)
If both x and y are odd, their sum is even, so
φ(x + y) = 0 = 1 +2 1 = φ(x) +2 φ(y)
Finally, if x is even and y is odd, then x + y is odd, and
φ(x + y) = 1 = 0 +2 1 = φ(x) +2 φ(y)
So φ is a homomorphism. It is not an isomorphism, which can be shown by using the definition
of an isomorphism or by observing that the set 2Z is the kernel of φ; if φ were an isomorphism,
its kernel would be trivial.
24. If f : G → H is a group homomorphism. Then
f (eG )eH = f (eG ) = f (eG eG ) = f (eG )f (eG )
Then by the cancellation law,
eH = f (eG )
25. First, note that since 0 must be in both ideals (they are both rings, so they must contain the additive
identity), we know the intersection is nonempty. Next, suppose a, b ∈ I ∩ J. To show that I ∩ J is an
ideal, we’ll use the ideal test:
(a) Since a is in both I and J, which are ideals, then for any r ∈ R we must have both ra and ar
in I and in J, so ra and ar are in I ∩ J.
(b) Since a − b must be in both I and J, then a − b must also be in I ∩ J.
Hence, I ∩ J is an ideal of R.
26. (a) The ideals hmi and hni contain, respectively, the multiples of m and n.
(b) The ideal hmi ∩ hni contains all of the integers that are multiples of both m and n, and as such
is generated by the least common multiple of these integers:
hmi ∩ hni = hlcm(m, n)i
There is one special case to consider: if either n or m is 0, then hmi ∩ hni = {0}, the trivial
ideal.
(c) First, we’ll assume both n and m are nonzero integers. In that case, every integer in the ideal
hmi + hni can be written in the form
mk + nl
for integers k and l. In other words, every integer in the ring hmi + hni is a multiple of the
greatest common factor of m and n, so
hmi + hni = hgcf(m, n)i
An interesting case is when m and n are relatively prime - then the sum of the ideals they generate
is the entire ring of integers.
Now, without loss of generality, suppose n = 0. Then h0i + hmi = hmi.
6
27. (⇒) Suppose R is an integral domain. If ab = ac and we know a 6= 0, then we can write
0 = ab − ac = a(b − c)
since rings have the distributive property. By definition of an integral domain, R has no zero divisors,
so we must have b − c = 0. Then b = c.
(⇐) This is simple to prove using the contrapositive of the statement (recall that the contrapositive
of a statement P → Q is Q → P , and that these statements are logically equivalent). Suppose R is
not an integral domain, so R contains elements x, y such that xy = 0 but neither x nor y is equal to
0. Since xy = 0, then xy = 0y. But if the cancellation law held, we’d be able to ’cancel’ y to obtain
x = 0, which counters our assumption.
28. (a) This would not be a ring homomorphism. As an example, phi(1) should map to 1, as homomorphisms preserve unity, but φ(1) = 4 6= 1.
(b) This would also not be a homomorphism, which we can again notice by seeing what happens to
0 and 1. Note that
φ(0, 1) = φ(1, 0) = 0
but
φ((0, 1) + (1, 0)) = φ(1, 1) = 1 6= 0 + 0 = φ(0, 1) + φ(1, 0)
(c) We finally have a homomorphism. Let f (x), g(x) be polynomials in R[x]. Then
φ(f (x) + g(x)) = φ((f + g)(x)) = (f + g)(1) = f (1) + g(1) = φ(f (x)) + φ(g(x))
φ(f (x)g(x)) = φ(f g(x)) = f g(1) = f (1)g(1) = φ(f (x))φ(g(x))
29. (a) The ring Zn for any n that is NOT prime is commutative and contains 1, but is not an integral
domain because it contains zero divisors. For instance, in Z6 , 2 · 3 = 0.
(b) The set of even integers 2Z does not contain the multiplicative identity, so it forms a commutative
ring without unity.
(c) The integers are an integral domain, since they form a commutative ring and do not contain zero
divisors. However, the nonzero integers do not form a group under multiplication (the inverse
property does not hold). So, they do not form a group.
(d) The sets of real, rational, and complex numbers each form fields.
(e) The set Zp for any prime p is a field with a finite number of elements. For example, Z7 is a field
with 7 elements.
(f) The ring of 2×2 matrices with real entries would work here. We’re not restricting to the matrices
with nonzero determinant, so we can’t assume they are invertible. Matrix multiplication is not
commutative, but there is an identity element in the set. Finally, the ring contains zero divisors,
since for example
1 0
0 0
0 0
=
0 0
1 0
0 0
30. (a) 2x2 + 2x, which has zeros 0 and 2 in Z3 .
(b) 1, which has no zeros in Z7 .
31. There’s always the trivial homomorphism:
φ(f (x)) = 0
Something more interesting would be the derivative, or a constant multiple of the function, such as
φ(f (x)) = 4f (x)
7