Here are SOLUTIONS to selected practice problems

```ÇANKAYA UNIVERSITY
Department of Mathematics and Computer Sciences
MCS 323 Abstract Algebra
Practice Problems-1
Gallian
(p.148)
7. Find all of the left cosets of H = {1, 11} in U (30).
8. Suppose that a has order 15. Find all of the left cosets of a5 in a.
SOLUTION:
5 5 10 a = a , a , e ; the left cosets are a5 , a a5 , a2 a5 , a3 a5 , and a4 a5 .
9. Let |a| = 30. How many cosets of a4 in a are there?
10. Let a and b be nonidentity elements of different orders in a group G of order 155. Prove
that the only subgroup of G that contains a and b is G itself.
SOLUTION:
By Lagrange’s Theorem, the orders of a and b are divisors of |G| = 155 = 5 • 31, and neither
is 1, since a, b = e. Let H ≤ G contain both a and b. If either of a, b has order 155, then
obviously |H| ≥ 155, so |H| = 155, so H = G. Otherwise the orders of a and b are 5 and 31 in
some order; but then 5 and 31 both divide |H|, so again |H| ≥ 155, so again H = G.
13. Let |G| = 60. What are the possible orders for the subgroups of G?
14. Suppose that K is a proper subgroup of H and H is a proper subgroup of G. If |K| = 42
and |G| = 420, what are the possible orders of H?
SOLUTION:
We need 42 dividing |H| and |H| dividing 420, with |H| =
42, 420. The multiples of 42 that
are divisors of 420 are 42 • 2 = 84 and 42 • 5 = 210.
15. Suppose that |G| = pq, where p and q are prime. Prove that every proper subgroup of
G is cyclic.
19. Suppose |G| = n and gcd (m, n) = 1. If g ∈ G and g m = e, prove that g = e
20. Suppose H, K are subgroups of G. If |H| = 12 and |K| = 35, find |H ∩ K|.
SOLUTION:
Note that there are many groups G for which the condition holds. We don’t need to investigate G. Instead, we notice that H ∩ K ≤ K and H ∩ K ≤ H. It tells us that the order
of H ∩ K divides the order of H, and the order of K. Since, gcd (|H| , |K|) = 1, we get that
|H ∩ K| = 1.
21. Suppose G is abelian of odd order. Show that the product of all of the elements of G is
the identity.
SOLUTION:
The order of any element must divide the order G, by Lagrange’s theorem; since G has odd
order then no element can have order 2. Hence for every x = e in G, x = x−1 . Thus we can
−1
−1
write the product of all the elements of G as ea1 a−1
1 a2 a2 · · · ak ak = e for some appropriate
choice of a1 , a2 , · · · , ak .
22. Suppose |G| > 1 and G has no proper nontrivial subgroups. Prove that |G| is prime.
SOLUTION:
Choose an x ∈ G \ {e}. Then x ≤ G; since G has
no proper, nontrivial subgroups, then,
x = G. Thus G is cyclic. If G is infinite, then x2 is a nontrivial proper subgroup of G, so
G must not be infinite. Thus G is a finite cyclic group. By the Fundamental Theorem of finite
cyclic groups, for each divisor k of |G| there is a subgroup of order k of |G|; thus G must not
have any divisors other than itself and 1. Hence |G| is prime, as desired.
23. Let |G| = 15. If G has only one subgroup of order 3 and only one of order 5, prove that
G is cyclic.
24. Let |G| = 25. Prove that G is cyclic or g 5 = e for all g ∈ G.
SOLUTION:
Suppose G is not cyclic. Choose any nonidentity g ∈ G. By Lagrange’s Theorem,
|g| | |G| = 25, so the order of g must 5 or 25. But we have assumed that G is not cyclic, so
not element can have order 25, so |g| = 5. This holds for any nonidentity g ∈ G, and certainly
e5 = e, so g 5 = e for all g ∈ G.
25. Let |G| = 33. What are the possible orders for the elements of G? Show that G must
have an element of order 3.
26. Let |G| = 8. Show that G must have an element of order 2.
27. Let G be finite and H ⊆ K ⊆ G be subgroups of G. Prove that |G : H| = |G : K| |K : H|.
28. Show that (Q, +) has no proper subgroup of finite index.
SOLUTION:
Suppose that H < Q has finite index |Q : H| = n. Choose any nonzero x ∈ Q. Consider the
left cosets H, x + H, 2x + H, · · · , nx + H. These are n + 1 cosets, so since H has index n in
Q, they cannot be distinct; two of them must be equal. Say, then, that ix + H = jx + H with
i < j, so (j − i) x + H = H so (j − i) x ∈ H. Now, 1 ≤ j − i ≤ n, so n! is a multiple of j − i,
so n!x ∈ H.
x
∈ H, so we have
Hence for every nonzero x ∈ Q, n!x ∈ H. Now for any nonzero x ∈ Q,
n!
1
(n!x) ∈ H; but this just says that every nonzero x ∈ Q is in H. Thus H = Q, contradicting
n!
the fact that H < Q. We have reached a contradiction, so there can be no H < Q with finite
index.
30. Prove that every subgroup of Dn of odd order is cyclic.
31. Let G = {(1) (12) (34) , (1234) (56) , (13) (24) , (1432) (56) , (56) (13) , (14) (23) , (24) (56)}.
a. Find the stabilizer of 1 and the orbit of 1.
b. Find the stabilizer of 3 and the orbit of 3.
c. Find the stabilizer of 5 and the orbit of 5.
33. Let |G| = pn where p is a prime. Prove that |Z (G)| =
pn−1 .
SOLUTION:
Suppose |Z (G)| = pn−1 and take any a ∈ G \ Z (G). Then CG (a) contains both a and Z (G),
so CG (a) is strictly larger than Z (G). That means that |Z (G)| = pn−1 divides |CG (a)| = pn ,
so CG (a) = G. This works for any a ∈ G (since also CG (a) = G for every a ∈ Z (G)), so G is
abelian. But then Z (G) = G, so |Z (G)| = pn , contradicting the fact that |Z (G)| = pn−1 .
There is a shorter proof.
pn
|G|
= n−1 = p. But then G/Z (G) has prime
If |Z (G)| = pn−1 then |G/Z (G)| =
|Z (G)|
p
order and is therefore cyclic, whence by the G/Z (G) theorem, G is abelian. Then Z (G) = G,
contradicting the fact that |Z (G)| < |G|.
34. Prove that a group of order 12 must have an element of order 2.
SOLUTION:
Let G be a group of order 12. Suppose that G has no element of order 2. If G has an element
g of order 4, then g 2 has order 2, so G has no element of order 4. Similarly, if some g has order
6 or 12, then g 3 or g 6 has order 2, so there are no elements of order 6 or 12. The order of
any element divides 12, so the only orders left are 1 and 3. Hence each of the 11 nonidentity
elements of G has order 3. But elements of order 3 come in pairs, since if |a| = 3, then a2 = 3,
so there cannot be exactly 11 of them. Hence G must have an element of order 2.
37. Show that in a group G of odd order, the equation x2 = a has a unique solution for all
a ∈ G.
SOLUTION:
Suppose |G| = 2k + 1 and x, y are both solutions of x2 = a. We have x = xe = xx2k+1 =
k+1
= y 2k+2 = yy 2k+1 = ye = y, so x = y. Hence the solution is unique..
x2k+2 = x2
(page 165)
2. Show that Z2 ⊕ Z2 ⊕ Z2 has seven subgroups of order 2.
5. Prove or disprove that Z ⊕ Z is cyclic.
SOLUTION:
Consider (a, b) for any a, b ∈ Z. Any element of (a, b) has a multiple of b in its second
coordinate, by which (a, b + 1) ∈ (a, b) unless b = 1, in which case (a + 1, b) ∈ (a, b) by
similar reasoning. Either way we have an element of Z ⊕ Z not in (a, b), so no (a, b) generates
Z ⊕ Z, so Z ⊕ Z is not cyclic.
6. Prove, by comparing orders of elements, that Z8 ⊕ Z2 is not isomorphic to Z4 ⊕ Z4 .
7. Prove that G1 ⊕ G2 is isomorphic to G2 ⊕ G1 .
8. Is Z3 ⊕ Z9 isomorphic to Z27 ? Why?
9. Is Z3 ⊕ Z5 isomorphic to Z15 ? Why?
10. How many elements of order 9 does Z3 ⊕ Z9 have?
11. How many elements of order 4 does Z4 ⊕ Z4 have? How many elements of order 4 does
Z8000000 ⊕ Z400000 have?
SOLUTION:
In both Z4 ⊕ Z4 and Z8000000 ⊕ Z400000 , |(a, b)| = 4 iff |a| = 4 and |b| = 1, 2, or 4, and |a| = 1
or 2 (we already counted the case that |a| = |b| = 4). By a theorem we proved in class, if d | n,
then the number of elements of order d in Zn is φ (d). Hence for the first case, we have φ (4) = 2
choices for a and φ (4) + φ (2) + φ (1) = 4 choices for b, giving us 2 · 4 = 8 choices in all. For
the second case, we have φ (4) = 2 choices for b and φ (2) + φ (1) = 2 choices for a, giving us
2 · 2 = 4 choices. Hence there are 8 + 4 = 12 elements of order 4 in each of the given groups.
12. The dihedral group Dn of order 2n, (n ≥ 3) has a subgroup of n rotations and a subgroup
of order 2. Explain why Dn cannot be isomorphic to the external direct product of two such
groups.
13. Prove that (C, +) ∼
= R ⊕ R.
∼
14. Suppose that G1 = G2 and H1 ∼
= H2 . Prove that G1 ⊕ H1 ∼
= G2 ⊕ H2 .
15. If G ⊕ H is cyclic, prove that G and H are cyclic.
16. Determine the number of elements of order 15 and the number of cyclic subgroups of
order 15 in Z30 ⊕ Z20 .
17. What is the order of any nonidentity element of Z3 ⊕ Z3 ⊕ Z3 ?
20. The group S3 ⊕ Z2 is isomorphic to one of the following groups: Z12 , Z6 ⊕ Z2 , A4 , D6 .
Determine which one by elimination.
21. Show that H = {(g, g) | g ∈ G} (called diagonal subgroup)is a subgroup of G ⊕ G.
22. Find a subgroup of Z4 ⊕ Z2 that is not of the form H ⊕ K, where H ≤ Z4 and K ≤ Z2 .
23. Find all subgroups of order 3 in Z9 ⊕ Z3 .
SOLUTION:
Subgroups of order 3 are cyclic, so we just need to find the elements of Z9 ⊕ Z3 that have
order 3. These are easily seen to be (3, 0) , (3, 1) , (3, 2) , (0, 1), and (0, 2).
24. Find all subgroups of order 4 in Z4 ⊕ Z4 .
25. What is the largest order of any element in Z30 ⊕ Z20 ?
26. How many elements of order 2 are in Z2000000 ⊕ Z4000000 ?
27. Find a subgroup of Z800 ⊕ Z200 that is isomorphic to Z2 ⊕ Z4 .
28. Find a subgroup of Z12 ⊕ Z4 ⊕ Z15 that is of order 9.
29. Prove that R∗ ⊕ R∗ ∼
C∗ .
=
34. Determine the number of cyclic subgroups of order 15 in Z90 ⊕ Z36 .
37. Express Aut (U (25)) in the form Zm ⊕ Zn .
38. Determine Aut (Z2 ⊕ Z2 ).
40. Is Z10 ⊕ Z12 ⊕ Z6 ∼
= Z60 ⊕ Z6 ⊕ Z2 ?
41. Is Z10 ⊕ Z12 ⊕ Z6 ∼
= Z15 ⊕ Z4 ⊕ Z12 ?
42. Find an isomorphism from Z12 to Z4 ⊕ Z3 .
43. How many isomorphisms are there from Z12 to Z4 ⊕ Z3 ?
44. Suppose φ : Z3 ⊕ Z5 −→ Z15 is an isomorphism and φ (2, 3) = 2. Find the element in
Z3 ⊕ Z5 that maps to 1.
47. Determine all cyclic groups that have exactly two generators.
50. Express U (165) as an external direct product of cyclic groups of the form Zn .
51. Express U (165) as an external direct product of U-groups in four different ways.
52. Determine how many elements of Aut (Z20 ) have order 4? How many have order 2?
53. Determine how many elements of Aut (Z720 ) have order 6?
54. Without doing any calculations in U (27), decide how many subgroups U (27) as.
55. What is the largest order of any element in U (900)?
56. Prove that U (55) ∼
= U (75).
57. Prove that U (144) ∼
=U (140).
2
58. Prove that U (n) = x2 | x ∈ U (n) is a proper subgroup of U (n).
60. Show that U (55)3 = x3 | x ∈ U (55) is U (55).
61. Find an integer n such that U (n) contains a subgroup isomorphic to Z5 ⊕ Z5 .
62. Find a subgroup of order 6 in U (700).
63. Show that there is a U -group containing a subgroup isomorphic to Z3 ⊕ Z3 .
64. Show that no U -group has order 14.
65. Show that there is a U -group containing a subgroup isomorphic to Z14 .
66. Show that no U -group is isomorphic to Z4 ⊕ Z4 .
67. Show that there is a U -group containing a subgroup isomorphic to Z4 ⊕ Z4 .
(p.191)
9. Let G = Z4 ⊕ U (4) , H = (2, 3), and K = (2, 1). Show that G/H ∼
= G/K. (This
shows that H ∼
= K does not imply that G/H ∼
= G/K.)
10. Prove that a factor group of a cyclic group is cyclic.
SOLUTION:
Let G = a. Then, we claim: G = H = aH. First, it should be clear that aH ∈ G/H (as
G/H = {gH | g ∈ G} modulo the equivalence class). This shows that aH ⊆ G/H.
Now, let wH ∈ G/H. Then, w ∈ G and so w = an for some n. This says that wH = (aH)n ,
i.e., wH ∈ aH. Hence, G/H = aH. Therefore GH is cyclic.
11. Let H G. If H and G/H are abelian, must G be abelian?
12. Prove that a factor group of an abelian group is abelian.
SOLUTION:
Let G be abelian and H ≤ G (so H G). Take aH, bH ∈ G/H. We have aHbH = abH =
baH = bHaH where the middle equality holds because G is abelian.
13. If H < G and a, b ∈ G, prove that (ab) H = a (bH).
SOLUTION:
(ab) H is the set of products of the form (ab) h where h ∈ H. Now (ab) h = a (bh) by
associativivty, so (ab) H is also the set of products of the form a (bh) f orh ∈ H; but this is just
a (bH)
14. What is the order 14 + 8 in the quotient Z24 / 8?
SOLUTION:
In this case, we have Z24 / 8 ∼
= Z8 . This is because a factor group of a cyclic group must be
cyclic, and all cyclic groups are isomorphic to Zn for some n.
Now, 14 + 8 = 6 + 8. Hence, we need to find the order of 6 in Z8 . We already have a
formula for this,
8
8
= = 4.
|6| =
gcd (6, 8)
2
A direct calculation will work equally well.
17. Let G = Z/ 20 and H = 4 / 20. List the elements of H and G/H.
18. What is the order of the factor group Z60 / 15?
SOLUTION:
We know that the order of a factor group is equal to the number of cosets.
|Z60 / 15| = |Z60 : 15| =
60
= 15.
4
19. What is the order of the factor group (Z10 ⊕ U (10)) / (2, 9)?
21. Prove that an abelian group of order 33 is cyclic.
22. Determine the order of Z ⊕ Z/ (2, 2). Is the group cyclic?
SOLUTION:
Notice that the set {(n, 0) | n ∈ Z} is an infinite subset of Z ⊕ Z/ (2, 2)", since (m, 0) +
(2, 2) = (n, 0) + (2, 2) iff n = m. Hence this factor group is of infinite order. We now show
that this group cannot be cyclic.
Claim. An infinite cyclic group cannot contain an element of finite order.
Proof. Suppose G is an infinite cyclic group, say G = a, and b ∈ G is an element of
finite order, say |b| = n. Then since a generates G b = a! + a +"#· · · + a\$k = ka for some
k ∈ Z, so that 0 = bn = nb = n (ka) = ank or in other words, a has finite order. But a
generates an infinite cyclic group, so this is a contradiction. This proves the claim. To show
that Z ⊕ Z/ (2, 2) is not cyclic, then, we need only to produce an element of finite order. Note
that (1, 1) + (2, 2) + (1, 1) + (2, 2) = (2, 2) + (2, 2) = (0, 0) + (2, 2). This completes the
proof.
23. Determine the order of Z ⊕ Z/ (4, 2). Is the group cyclic?
24. The group Z4 ⊕ Z12 / (2, 2) is isomorphic to one of Z8 , Z4 ⊕ Z2 , or Z2 ⊕ Z2 ⊕ Z2 .
Determine which one by elimination.
25. Let G = U (32) and H = {1, 31}. The group G/H is isomorphic to one ofZ8 , Z4 ⊕ Z2 ,
or Z2 ⊕ Z2 ⊕ Z2 . Determine which one by elimination.
27. Let G = U (16), H = {1, 15}, and K = {1, 9}. Are H and K isomorphic? Are G/H and
G/K isomorphic?
28. Let G = Z4 ⊕ Z4 , H = {(0, 0) , (2, 0) , (0, 2) , (2, 2)}, and K = (1, 2). Is G/H isomorphic
to Z4 or Z2 ⊕ Z2 ? Is G/K isomorphic
to Z4 or Z2 ⊕ Z2 ? 29. Let G = GL (2, R) H = A ∈ G | det (A) = 3k , k ∈ Z . Prove that H G.
30. Express U (165) as an internal direct product of proper subgroups in four different ways.
34. In Z, let H= 5 and K = 7.Prove that Z = HK. Does Z = H× K?
35. Let G = 3a 6b 10c | a, b, c ∈ Z under multiplication and H = 3a 6b 12c | a, b, c ∈ Z
under multiplication. Prove that G = 3 × 6 × 10 whereas G = 3 × 6 × 12.
37. Let G be finite and H G. Prove that the order of gH must divide |g|.
38. Let H G, a ∈ G. If aH has order 3 in G/H and |H| = 10. What are possibilities for
|a|
SOLUTION:
First, aH has order 3 in G/H. Hence G = H ∪ aH ∪ a2 H has 30 elements and is a subgroup
of G: if we take x, y ∈ G then x = ai h1 , y = aj h2 for some i, j and h1 , h2 ∈ H. Then
−1
xy −1 = ai h1 aj h2
=
=
=
=
−j
ai h1 h−1
2 a
ai ha−j [for h = h1 h−1
2 ∈ H]
i −j a a h [for some h ∈ H, since H G]
ai−j h
and this last is an element of G . Hence a is an element of a group of order 30, so |a| divides
30.
Now |a| must be a nultiple of 3. If not, say |a| = 3k + i where i = 1 or 2. Then H = eH =
k
3k+i
H = (aH)3k+i = (aH)3 (aH)i = H k (aH)i = H (aH)i = (aH)i = ai H; so H = ai H for
a
i = 1, 2, so aH does not have order 3 in G/H, a contradiction.
Thus |a| is a multiple of 3 that divides 30. The possibilities for |a| are therefore 3, 6, 15, and
30. We can give examples of all 4 possibilities: In G = Z30 , let H = 3. Then for a = 10, 5, 2,
and 1, aH has order 3 and a has order 3, 6, 15, and 30, respectively.
39. If H G, then prove that CG (H) G.
40. Suppose G is abelian and H < G. If every element of H is a square and every element
of G/H is a square, then prove that every element of G is a square.
41. Show, by an example, that in a factor group G/H it can happen that aH = bH but
|a| =
|b|.
SOLUTION:
Take G = Z6 , H = {0, 3} , a = 1, and b = 4.
43. Suppose G is nonabelian with |G| = p3 (where p is a prime) and Z (G) = {e}. Prove
that |Z (G)| = p.
SOLUTION:
Use "G/Z Theorem."
44. If |G| = pq, where p, q are primes, prove that |Z (G)| = 1 or pq.
45. Let N G, H ≤ G and let N ≤ H. Prove that H/N G/N iff H G.
46. Let G be abelian and let H = {x ∈ G : |x| < ∞}. Prove that every nonidentity element
of G/H has infinite order.
47. Determine all subgroups of R∗ that have finite index.
50. Show that the intersection of two normal subgroups of G is a normal subgroup of G.
51. Let N G, H ≤ G. Prove that N H ≤ G. Give an example to show that N H need not
be a normal subgroup of G if neither N nor H is normal.
52. If N, M G prove that NM G.
53. Let N G. If N is cyclic, prove that every subgroup of N is normal in G.
SOLUTION:
r
m
k
Let N = a , H = ak , and x ∈ G. Then x ak x−1 = xam x−1 = (ar )k = ak ∈ H.
55. Let G be finite, H G and let x ∈ G. If gcd (|x| , |G/H|) = 1, show that x ∈ H.
SOLUTION:
gcd (|x| , |G/H|) = 1 implies gcd (|xH| , |G/H|) = 1. But |xH| divides |G/H|. Thus |xH| = 1
and therefore xH = H =⇒ x ∈ H. 56. Let G = x−1 y −1 xy | x, y ∈ G .
a) Prove that G is normal in G.
b) Prove that G/G is abelian.
c) If G/N is abelian, prove that G ≤ N .
d) Prove that if H ≤ G and G ≤ N , then H is normal in G.
57. If N G and |G/N | = m, show that xm ∈ N .
SOLUTION:
Let G be a group, with N a normal subgroup such that |G/N | = m. For all x ∈ G, we have
that xN ∈ G/N . Since G/N is itself a group of order m, we know that (xN)m = e. That is
e = xm N = N . In general, for a normal subgroup N, aN = N a if and only if a ∈ N. Therefore,
we can conclude that xm ∈ N .
58. Suppose G has a subgroup of order n. Prove that the intersection of all subgroups of G
of order n is a normal subgroup of G.
59. If G is nonabelian, show that Aut (G) is not cyclic.
SOLUTION:
We know that for any group G, we have G/Z (G) ∼
= Inn (G) ≤ Aut (G). So if Aut (G) were
cyclic, then Inn (G) would be cyclic. Thus G/Z (G) would have to be cyclic. But in this case G
would have to be abelian. This is contradiction to the given information that G is nonabelian.
60. Let |G| = pn m, where p is prime and gcd (p, m) = 1. Suppose H G with |G| = pk . If
K < G, |K| = pk , show that K ⊆ H.
61. Suppose G is finite and H G. If G/H has an element of order n, show that G has an
element of order n.
SOLUTION:
Say |gH| = n. Then |g| = nt by Exercise 37, and g t = n. For the second part, consider
Z/ k.
62. Recall that a subgroup N of a group G is called characteristic if φ (N) = N (we denote
this by N char G)for all φ ∈ Aut (G). If N char G, show that N G.
64. Show that S4 has a unique subgroup of order 12.
65. If |G| = 30 and |Z (G)| = 5, what is the structure of G/Z (G)?
SOLUTION:
30
We know that |G/Z (G)| =
= 6 = 2 · p, where p = 3 is a prime greater than 2 We also
5
know that any group of order2p, where p ia prime greater than 2 is isomorphic to Z2p or the
dihedral group Dp of order 2p. In the former case, since Z2p is cyclic, then G/Z (G) is cyclic,
and so G must be abelian by a theorem we proved in class, but this cannot be the case, because
in this case by the hypothesis, we have G = Z (G). In the latter case, we have G/Z (G) ∼
= Dp .
66. If H G and |H| = 2, prove that H ⊆ Z (G).
SOLUTION:
First, H contains the identity, e, and one other element, call it g. Since H is normal, for any
given a ∈ G, aH = Ha. Since ae = ea = a for all a, it must therefore be true that ag = ga for
all a, or else H would not be normal. So, all elements of H (namely e and g) commute with
all elements of G, so that H is a subset of Z (G).
67. Prove that A5 cannot have a normal subgroup of order 2.
SOLUTION:
Suppose that A5 has a normal subgroup of order 2, call it H. Then H is a subset of Z (A5 )
by 9.66. Recall from problem 5.46 that the center of Sn is trivial for all n ≥ 3. An argument
similar to the one presented there shows that A5 , a subgroup of S5 , also has a trivial center.
Since Z (A5 ) is trivial, it can have no subset of order 2, contradicting the assumption that A5
has a normal subgroup of order 2.
68. Let G be finite and let |H| be odd with |G : H| = 2. Show that the product of all the
elements of G (taken in any order) cannot belong
to H.
69. Let G be a group. If H = g 2 | g ∈ G is a subgroup of G, prove that it is a normal
subgroup of G.
SOLUTION:
2
Observe that xg 2 x−1 = xgx−1 .
70. Suppose H G. If |H| = 4 and gH has order 3 in G/H, find a subgroup of order 12
in G.
SOLUTION:
We begin with the following general result.
Proposition. Let G be a group and H G. For any g ∈ G the set
gi H
K=
i∈Z
is a subgroup of G.
Proof 1: We use the one-step subgroup test. We begin by noting that K = ∅ since H ⊂ K
The conclusion of the problem now follows easily. Since gH has order 3, the cosets H, gH
and g 2 H are distinct, and any other coset of the form g i H is one of these. Therefore
g i H = H ∪ gH ∪ g 2 H
i∈Z
and the latter set contains exactly 12 elements since |H| = 4. The proposition tells us this set
is a subgroup of G, so we’re finished.
71. Prove that A4 ⊕ Z3 has no subgroup of order 18.
(p.210)
9. Prove that the mapping from G ⊕ H −→ G given by (g, h) −→ g is a homomorphism.
What is the kernel? This mapping is called the projection of G ⊕ H onto G.
10. Let G be a subgroup of some dihedral group. For each x ∈ G, define
+1 if x is a rotation,
φ (x) =
−1 if x is reflection.
Prove that φ is a homomorphism from G to the multiplicative group {+1, −1}.
11. Prove that (Z ⊕ Z) / ((a, 0) , (0, b)) is isomorphic to Za ⊕ Zb .
SOLUTION:
Consider φ : Z ⊕ Z −→ Za ⊕ Zb given by φ (x, y) = (x mod a, y mod b). Then φ is a homomorphism from Z⊕Z onto Za ⊕Zb with kernel ker φ = {(x, y) ∈ Z ⊕ Z | φ (x, y) = (x mod a, y mod b) = (0, 0)}
(a, 0) , (0, b). By the first ismorphism theorem, we have (Z ⊕ Z) / ((a, 0) , (0, b)) ∼
= Za ⊕ Zb .
∼
12. Suppose that k is a divisor of n. Prove that Zn / k = Zk .
SOLUTION:
Define the homomorphism f : Zn −→ Zk as f (a mod n) = a mod k. First we check that this
is well defined. Let x, y ∈ Zn ∼
= Z/nZ be equal, i.e., x ≡ y (mod n), i.e., n | (x − y). But we are
given that k | n, so k | (x − y) means that x ≡ y (mod k), or x = y in Zk ∼
= Z/kZ. This
means that f (x + nZ) = x + kZ = y + kZ = f (y + nZ), so the map is well-defined.
Note that f is onto, since if a + kZ ∈ Z/kZ, then there is an equivalence class a + nZ ∈ Z/nZ
for which f (a + nZ) = a + kZ.
f is a homomorphism, since f ((a + nZ) + (b + nZ)) = f ((a + b) + nZ) = (a + b) + kZ =
(a + kZ) + (b + kZ) = f (a + nZ) + f (b + nZ) .
ker f = {a + nZ | f (a + nZ) = 0 + kZ}
= {a + nZ | a ≡ 0 (mod k)}
= k
Hence, by the First Isomorphism Theorem, Zn / k ∼
= Zk .
13. Prove that (A ⊕ B) / (A ⊕ {e}) ∼
B.
=
SOLUTION:
The mapping φ : A ⊕ B −→ B which is given by φ (a, b) = b is a homomorphism form A ⊕ B
onto B with kernel
ker φ =
=
=
=
{(a, b) ∈ A ⊕ B | φ (a, b) = eB }
{(a, b) ∈ A ⊕ B | b = e}
{(a, e) | a ∈ A}
A ⊕ {e}
So, by the first isomorphism theorem, we have (A ⊕ B) / ker φ ∼
= Im φ, that is, we have
(A ⊕ B) / (A ⊕ {e}) ∼
= B.
14. Explain why the correspondence x → 3x from Z12 to Z10 is not a homomorphism.
SOLUTION:
We show that φ (x) = 3x is not a homomorphism from Z12 to Z10 . Consider x = 6 and y = 7,
then x + y = 13 ≡ 1 (mod 12) φ (1) = 3 but φ (x) + φ (y) = 3 (6) + 3 (7) = 18 + 21 = 39 ≡
9 (mod 10).
15. Suppose φ : Z30 −→ Z30 is a homomorphism and ker (φ) = {0, 10, 20}. If φ (23) = 9,
determine φ−1 (9).
SOLUTION:
We have
φ−1 (9) =
=
=
=
23 + ker φ
23 + {0, 10, 20}
{23 + 0, 23 + 10, 23 + 20}
{3, 13, 23}
16. Prove that there is no homomorphism from Z8 ⊕ Z2 onto Z4 ⊕ Z4 .
SOLUTION:
Let us assume otherwise that there exists an onto homomorphism φ : Z8 ⊕ Z2 −→ Z4 ⊕ Z4 .
Note that we have |Z8 ⊕ Z2 | = 16 and |Z4 ⊕ Z4 | = 16. Since these group have the same number
of elements and is onto, has to be one-to-one. We assumed that φ is a homomorphism. So it
is operation-preserving. This means that φ is an isomorphism. By a theorem, we know that
|φ (g)| = |g| for every g ∈ Z8 ⊕ Z2 . Let us consider the order of the element (1, 0) ∈ Z8 ⊕ Z2 .
Since we know that Z8 = 1, we get |(1, 0)| = 8. But note that if (g1 , g2 ) is an element of
Z4 ⊕ Z4 , we have |(g1 , g2 )| = lcm (|g1 | , |g2 |) = 4 because |g1 | and |g2 | can only be 1, 2, 4. In
other words, Z4 ⊕Z4 doesn’t has an element of order 8. On the contrary, Z8 ⊕Z2 has an element
of order 8. This contradicts with one of the theorems we proved in class. Hence there is no
isomorphism between Z8 ⊕ Z2 and Z4 ⊕ Z4 .
17. Prove that there is no homomorphism from Z16 ⊕ Z2 onto Z4 ⊕ Z4 .
SOLUTION:
Suppose φ is such a homomorphism. By a theorem we proved in class, we have ker φ =
(8, 1) , (0, 1) or ker φ = (8, 1). In these cases, (1, 0) + ker φ has order either 16 or 8. So
Z4 ⊕ Z4 .
(Z16 ⊕ Z2 ) / ker φ ∼
=
18.
a) Can there be a homomorphism from Z4 ⊕ Z4 onto Z8 ?
b) Can there be a homomorphism from Z16 onto Z2 ⊕ Z2 ?
SOLUTION:
a.
There cannot be a homomorphism from Z4 ⊕ Z4 onto Z8 , because there exists an
element of order 8 in Z8 , namely 1, and there are no elements in Z4 ⊕ Z4 which have order
divisible by 8. This is the same as saying there are no elements in Z4 ⊕ Z4 which map to 1, and
thus any homomorphism from Z4 ⊕ Z4 onto Z8 cannot be surjective.
b.
There cannot be an onto homomorphism from Z16 onto Z2 ⊕ Z2 since the homomorphic
image of a cyclic group must be cyclic and Z2 ⊕ Z2 is not a cyclic group.
19. Suppose that there is a homomorphism φ from Z17 to some group and that φ is not
one-to-one. Determine φ.
SOLUTION:
Theorem 10.1 - property 4, implies, as ker φ ≤ Z17 that ker φ = {0}, since φ is not one-to-one.
Hence, as |ker φ| divides 17 by Lagrange, |ker φ| = 17, so ker φ = Z17 . Hence φ is the trivial
map, mapping all of Z17 to the identity.
20. How many homomorphisms are there from Z20 onto Z8 ? How many are there to Z8 ?
SOLUTION:
By a theorem a homomorphism φ : Z20 −→ Z8 is determined by the value of φ (1), since Z20
is cyclic. Indeed, if φ (1) = a, then for any n ∈ Z20 ,
φ (n) = φ 1 + 1 +· · · + 1 = φ (1) + φ (1) + · · · + φ (1) = a + a + · · · + a.
We know that |a| divides 20 which is the order of 1 ∈ Z20 . We know that Im (φ) ≤ Z8 , and
so by Lagrange’s theorem, |a| divides the order of |Im (φ)|, so |a| | 8. Hence |a| divides both 8
and 20, and so there are 3 possibilities for |a|, namely 1, 2, 4. This gives the four possibilities
a = 0, 2, 4, 6 ∈ Z8 , which already means that there are no homomorphisms from Z20 onto Z8 ,
since for any of these a’s a + a + · · · + a will never be odd. On the other hand, each of the 4
possibilities gives rise to a homomorphism, by inspection of the following explicit formulas:
21. If φ is a homomorphism from Z30 onto a group of order 5, determine ker φ.
SOLUTION:
The order of ker φ is |ker φ| = 5.
22. Suppose φ : G −→ G is an epimorphism and that G = 8. Prove that φ has an element
of order 8. Generalize.
SOLUTION:
It is given that G has an element of order 8. Let us call this element g. So we get |g| = 8.
Since φ is an onto homomorphism, there is an element g ∈ G so that φ (g) = g. By a theorem
we know that |φ (g)| = |g| divides |g|. This means 8 divides |g| because |g| = 8. So we obtain
that |g| = 8t for some positive integer t. Then if we consider the element g t in G, we get
t 8
g
= g t · g t · · · · · g t (8 copies of t)
= g 8t
= e.
Thus G has an element g t of order 8. As a generalization, If G is finite and G has an element
of order n, then so does G.
23. Suppose φ : Z36 −→ H is a homomorphism with |H| = 24.
a. Determine the possible homomorphic images.
b. For each image in part a., determine the corresponding ker φ.
24. Suppose φ : Z50 −→ Z15 is a homomorphism with φ (7) = 6.
a. Determine φ (x) .
b. Determine Im (φ).
c. Determine ker φ.
d. Determine φ−1 (3).
SOLUTION:
a.
Since a homomorphism between cyclic groups is determined by where we send 1, we have to do
a little arithmetic to determine φ (x). We are given that φ (7) = 6 but φ (7 + 7 + 7 + 7 + 7 + 7 + 7) =
φ (49) = φ (−1) = 42 = 12 and then φ (1 + (−1)) = φ (0) = 0 so it follows that φ (−1) + φ (1) =
0. This says that φ (1) = 3. Therefore φ (x) = 3x.
b.
The image of φ is all multiples of 3, or Im (φ) = 3.
c.
The kernel of φ is all multiples of 5, or ker φ = 5.
d.
The inverse image of 3 is the set {x ∈ Z50 | φ (x) = φ (1)}. So φ−1 (3) = 1 + ker φ.
25. How many onto homomorphisms are there from φ : Z20 −→ Z10 . How many to Z10 ?
SOLUTION:
There are 4 onto homomorphism φ : Z20 −→ Z10 . There are 10 homomorphism φ : Z20 −→
Z10 .
26. Determine all homomorphisms from φ : Z4 −→ Z2 ⊕ Z2 .
SOLUTION:
We know that Z4 = 1. Let φ : Z4 −→ Z2 ⊕Z2 be a homomorphism between Z4 and Z2 ⊕Z2 .
STEP 1: Find the possible orders for φ (1): We know that |1| = 4 in Z4 . By a theorem, |φ (1)|
must be a divisor of 4. We know that |(g1 , g2 )| = lcm (|g1 | , |g2 |) for every (g1 , g2 ) ∈ Z2 ⊕ Z2 .
Since we have |g1 | = 1 or 2, we get that |(g1 , g2 )| = 1 or 2. So |φ (1)| = 1 or 2.
STEP 2: Find all the elements in Z2 ⊕ Z2 of order 1 or 2: We have |(0, 0)| = 1, |(1, 0)| =
2, |(0, 1)| = 2, |(1, 1)| = 2. So, (0, 0) , (1, 0) , (0, 1) , (1, 1) are possible images for 1 in Z4 .
STEP 3: Determine all homomorphisms: the image of 1 determines the homomorphism
between Z4 and Z2 ⊕ Z2 . If we map
1 −→ (0, 0)
we get a homomorphism. Let us call this φ1 . If we map
1 −→ (1, 0)
we get a homomorphism. Let us call this φ2 . If we map
1 −→ (0, 1)
we get a homomorphism. Let us call this φ3 . If we map
1 −→ (1, 1)
we get a homomorphism. Let us call this φ4 . Therefore, there are 4 homomorphisms {φ1 , φ2 , φ3 , φ4 }
from Z to from Z2 ⊕ Z2 .
27. Determine all homomorphisms from Zn to itself.
SOLUTION:
For each k with 0 ≤ k ≤ n−1, the mapping φ : 1 −→ φ (1) = k determines a homomorphism.
28. Suppose φ : S4 −→ Z2 is a homomorphism. Determine ker φ. Determine all homomorphisms from S4 to Z2 .
29. Suppose G is finite and φ : G −→ Z10 is a homomorphism. Prove that G has normal
subgroups of indexes 2 and 5.
SOLUTION:
By the First Isomorphism Theorem, we have G/ ker φ ∼
= Im φ ≤ Z10 . So the possibilities for
the order of G/ ker φ are |G/ ker φ| = 1, 2, 5, or 10.
30. Suppose φ : G −→ Z6 ⊕ Z2 is an epimorphism and |ker φ| = 5.
31. Suppose φ : U (30) −→ U (30) is a homomorphism with ker φ = {1, 11}. If φ (7) = 7,
find φ−1 (7).
SOLUTION:
We have
φ−1 (7) = 7 ker (φ) = 7 {1, 11} = {7, 77 mod 30} = {7, 17}.
32. Find a homomorphism φ : U (30) −→ U (30) with ker φ = {1, 11} and φ (7) = 7.
SOLUTION:
First U (30) = {1, 7, 11, 13, 17, 19, 23, 29}. Then φ−1 (1) = ker φ = {1, 11} and φ−1 (7) =
{7, 17} . Next, what is φ−1 (11)? We know that φ−1 (11) ∩ {1, 7, 11, 17} = ∅. We know that
19has order 2 in U (30). The orders of order untouched elements are |13| = 4 , |17| = 4, |19| =
2, |23| = 4, |29| = 2. Since the order of 19 is not divisible by 4, there are two choices for φ (19).
Either φ (19) = 19, or φ (19) = 29. In the first case,φ−1 (19) = 19 ker φ = 19 {1, 11} = {19, 29}
and so in the second case φ (19) = 29, we have φ−1 (29) = 29 ker φ = 29 {1, 11} = {29, 19}
33. Suppose φ : U (40) −→ U (40) is a homomorphism with ker φ = {1, 9, 17, 33}. If
φ (11) = 11, find φ−1 (11).
SOLUTION:
We have
φ−1 (11) = 11 ker φ
= 11 {1, 9, 17, 33}
= {11, 19, 27, 3}
34. Find a homomorphism φ : U (40) −→ U (40) with ker φ = {1, 9, 17, 33} and φ (11) = 11.
SOLUTION:
It is possible that nothing maps to 11. However, suppose that φ (a) = 11. Then any element
of a ker φ = {a, 19a, 17a, 33a} also maps to 11 (where the multiplication is in U (40).
35. Prove that the mapping
φ : Z ⊕ Z −→ Z
(a, b) −→ φ (a, b) = a − b
is a homomorphism. Determine ker φ. Describe φ−1 (3).
SOLUTION:
Since
φ ((a, b) + (c, d)) =
=
=
=
it follows that φ is a homomorphism.
φ ((a + c, b + d))
(a + c) − (b + d)
a−b+c−d
φ (a, b) + φ (c, d) , ∀ (a, b) , (c, d) in Z ⊕ Z
Next for the kernel, we have
ker φ = {(a, b) | φ (a, b) = 0}
= {(a, b) | a − b = 0}
= {(a, a) | a ∈ Z} .
Finallay, we have
φ−1 (3) =
=
=
=
{(a, b) | φ (a, b) = 3}
{(a, b) | a − b = 3}
{(a, b) | a = b + 3}
{(a + 3, a) | a ∈ Z} .
36. Suppose φ : Z ⊕ Z −→ G is a homomorphism such that φ (3, 2) = a and φ (2, 1) = b.
Determine φ (4, 4) in terms of a and b. Assume that the operation of G is addition.
SOLUTION:
The whole point here is that every element of Z ⊕ Z can be written as a linear combination of
(3, 2) and (2, 1). This is because, given any (u, v) ∈ Z⊕Z, the equation x (3, 2)+y (2, 1) = (u, v)
is the same as the vector equation
3
2
u
x
+y
=
2
1
v
which is the same as the matrix equation
3 2
x
u
=
2 1
y
v
and the latter has the solution
−1 u
x
−1 2
3 2
u
−u + 2v
=
=
=
v
y
2 −3
2 1
v
2u − 3v
which is the vector with integer entries since u, v ∈ Z. From this, it follows that
φ ((u, v)) = φ (x (3, 2) + y (2, 1)) = xφ (3, 2) + yφ (2, 1) = (−u + 2v) a + (2u − 3v) b
In particular,
φ (4, 4) = (−4 + 8)a + (8 − 12)b = 4a − 4b.
37. Prove that the mapping
φ
: C∗ −→ C∗
x −→ φ (x) = x6
is a homomorphism. What is ker φ?
SOLUTION:
Since φ (xy) = (xy)6 = x6 y 6 = φ (x) φ (y) , ∀x, y ∈ C∗ , φ is a homomorphism. Next
ker φ = {x ∈ C∗ | φ (x) = 1}
= x ∈ C∗ | x6 = 1
π
π
= eiπ/6 = cos + i sin
6
6
52. Let α, β be homomorphisms from G to G and let H = {g ∈ G | α (g) = β (g)}. Prove or
disprove that H is a subgroup of G.
SOLUTION:
We will use the one-step subgroup test to prove that H is indeed a subgroup of G. First of
all, H = ∅; since α (e) = e = β (e) implies that e ∈ H. Now, if a, b ∈ H then
α ab−1 = α (a) α b−1 = α (a) α (b)−1 = β (a) β (b)−1 = β (a) β b−1 = β ab−1
implying that ab−1 ∈ H. Therefore H is a subgroup of G.
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