Selected Solutions to
Abstract Algebra (Dummit and Foote)
Henry Liu
October 29, 2013
Contents
1 Introduction to Groups
1.1 Basic Axioms and Examples . . . .
1.2 Dihedral Groups . . . . . . . . . .
1.3 Symmetric Groups . . . . . . . . .
1.4 Matrix Groups . . . . . . . . . . .
1.5 The Quaternion Group . . . . . . .
1.6 Homomorphisms and Isomorphisms
1.7 Group Actions . . . . . . . . . . .
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2
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2 Subgroups
2.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Chapter 1
Introduction to Groups
1.1
Basic Axioms and Examples
8. (a) Let a, b ∈ G. Then an = bn = 1, hence (ab)n = an bn = 1, and G is closed under multiplication.
Furthermore, G has the multiplicative identity 1, since 1n = 1. Finally, if a ∈ G, then aan−1 =
2
an = 1, but an−1 ∈ G since (an−1 )n = an /an = 1n /1 = 1, hence any element in G has an inverse.
By the definition of a group, G is a group under multiplication.
(b) Note that 1 ∈ G, since 1n = 1. But 1 + 1 = 2 6∈ G, since 2n 6= 1 for any n ∈ Z+ . Thus G is not
closed under addition and cannot be a group.
14.
• 1̄ has order 1, since it itself is the multiplicative identity.
¯ 2 = 1̄.
• 2̄ has order 2, since (−1)
¯ 17,
¯ 13,
¯ 29,
¯ 1̄.
• 5̄ has order 6, since the sequence of powers goes 5̄, 25,
¯ has order 3, by the same method of powers.
• 13
¯ 3 = 1̄.
¯ has order 6, either by finding powers or noting that −13
¯ 6 = (−1)
¯ 2 13
• −13
¯ has order 2, since 17
¯ 2 = 1̄.
• 17
20. Suppose x has order n, i.e. xn = 1 and xk 6= 1 for 0 ≤ k < n. Then (x−1 )k = x−k = (xk )−1 6= 1 for 0 ≤
k < n (this rearrangement of exponents is possible by a previous question), but (x−1 )n = (xn )−1 = 1,
hence x−1 has order n as well.
25. Since multiplicative inverses are unique, if x2 = 1, then x = x−1 . Then for a, b ∈ G, we have
(ab)2 = abab = a−1 b−1 ab = 1, and multiplying both sides by a and b gives ab = ba. Hence G is abelian.
27. Note that xn x−n = 1, and x−n ∈ {xn | n ∈ Z, hence that set is closed under inverses, and clearly
closed under the group operation. Hence by a previous problem, or by noting that the group operation
retains its associativity under a restriction and that the identity element is in the set, we have that the
given set is a subgroup.
30. By the definition of a direct product of groups, (a, 1)(1, b) = (a, b) = (1, b)(a, 1). If 1 = (a, b)n =
(a, 1)n (1, b)n = (an , 1)(1, bn ), then we must have an = bn = 1. This will happen for the smallest
x, y ∈ Z+ such that n = x|a| = y|b|, which is precisely the definition of n = lcm(|a|, |b|).
32. Suppose xj = xk for some 0 ≤ j < k ≤ n − 1. Then x−j xj = 1 = xk−j . But then x is of order
k − j < n, yet we are given that x is of order n. By contradiction, the elements xj for 0 ≤ j ≤ n − 1
are distinct.
2
1.2
Dihedral Groups
3. Note that an element of D2n which is not a power of r is of the form srk , for 0 ≤ k < n. If k = 0, then
s2 = 1, so suppose k > 0. Then we have
(srk )2 = (srk )(srk ) = srk−1 rsrk = srk−1 sr−1 rk = srk−2 sr−2 rk = . . . = ssr−k rk = 1
Hence srk has order 2. Note that s, r ∈ hs, sri since r = s · sr, hence D2n is also generated by s, sr.
7. From the relation a2 = 1, we get s2 = 1, from (ab)n = 1, we get (ssr)n = rn = 1, and from b2 = 1
we get srsr = 1 =⇒ rsr = s =⇒ rs = sr−1 . Conversely, from s2 = 1, we get a2 = 1, from
rn = 1, we get (ssr)n = (ab)n = 1, and from rs = sr−1 we get rsr = s =⇒ srsr = b2 = 1. Hence
a2 = b2 = (ab)n = 1 is also a presentation for D2n .
12. Note that in a dodecahedron, there are 20 vertices, each of which is connected to three others. Then
there are 20 possible locations for a selected vertex to be sent to, and a second vertex connected to the
selected vertex may go to any of three locations. Hence |G| = 20 · 3 = 60.
¯ 1.
15. Z/nZ has the presentation hx | xn = 1i, since its distinct elements are 0̄, 1̄, ..., n −
17. (a) Recall that every element of X2n can be written as xa y b , and that x3 = 1. Then 0 ≤ a < 3, and
0 ≤ b < 2, and since x and y are distinct, all of these elements xa y b must be distinct as well.
Hence |X2n | = 6. In addition, note that when x3 = 1, we have x2 = x−1 , hence the relation
xy = yx2 becomes xy = yx−1 , which makes X2n the same as the dihedral group D6 .
(b) If (3, n) = 1, then n = 3k + 1 for some k. Then 1 = x3k+1 = (x3 )k x = x. Hence X2n is only
composed of the elements 1 and y, and therefore has order 2.
18. (a) Since v 3 = 1, multiplying by its inverse gives v 2 = v −1
(b) Note that v 2 u3 v = (v 2 u2 )(uv) = (uv)(v 2 u2 ) = u3 . But v 2 = v −1 , so v −1 u3 v = u3 , and multiplying
by v on both sides gives u3 v = vu3 . Hence v commutes with u3 .
(c) We have u9 = u4 u4 u = u, hence vu = vu9 = vu3 u3 u3 = u3 u3 u3 v = u9 v = uv.
(d) uv = v 2 u2 as a relation, but uv = vu, hence v 2 u2 = vu, and multiplying by u−1 v −1 gives
vu = uv = 1.
(e) Since u4 = v 3 = 1, we have 1 = u4 v 3 = uu3 v 3 = u(uv)3 = u because u and v commute. Then
uv = v = 1, and therefore Y = 1.
1.3
1.
Symmetric Groups
• σ = (1 3 5)(2 4) and τ = (1 5)(2 3).
• From the above, σ 2 = (1 5 3) and τ 2 = 1.
• Again from the above, στ = (1 2 4 3) and τ σ = (2 5 3 4).
• Finally, since τ 2 = 1, we have τ 2 σ = σ.
8. Suppose SΩ has finite order. Then there is an element of SΩ that maps 1 to a maximal value n. But
that implies any map that maps 1 to a larger value is not in SΩ , contradicting the definition of SΩ .
Hence SΩ has infinite order.
10. Since σ(ak ) = ak+1 , by induction we have σ i (ak ) = ak+i wrapped mod m because σ(am ) = a1 . Then
it takes at least m applications of σ to return 1 to its original position, but at most m to return any
element to its original position since σ m = 1. Hence |σ| = m.
3
11. Suppose (i, m) = 1. Then σ i (n) = n + i wrapped mod m, but ki mod m 6= 0 for 0 < k < m, hence
it takes m applications of σ i until n returns to its original position, and thus σ i is an m-cycle. If
(i, m) 6= 1, however, there is some 0 < k < m such that ki mod m = 0, and an element n can be sent
back to its original position in k applications of σ i , hence σ i is not an m-cycle.
Q
14. Suppose σ has cycle decomposition σi where σi ∈ Sn commute and each have order p prime. Then
Y p Y
Y
σp =
1=1
σi =
σip =
Since σik 6= 1 for k < p and the cycles in a cycle decomposition are disjoint, σ k 6= 1, and hence |σ| = p.
Now suppose σ ∈ Sn has order p prime. Suppose it is not the product of commuting p-cycles, i.e. it
can be written as σ = τ ω where τ is a q-cycle with q 6= p. Then τ p is a q-cycle as well, by problem 11,
hence τ p 6= 1. Then σ p = τ p ω p 6= 1. By contradiction, σ is the product of commuting p-cycles.
For a counterexample to this when p is not prime, consider (1 2 3)(4 5) ∈ S6 . This has order 6, but is
not the product of commuting 6-cycles.
15. Recall that the cycles in a cycle
Q decomposition are disjoint and therefore commutative. Then for σ ∈ Sn
with cycle decomposition σi , we have
Y k Y
σi =
σik
σk =
If we want σ k = 1, then we want each σik = 1. Let the order of σi be pi . Then k = ci pi for some ci ’s,
but the smallest such k is precisely the lcm of all the pi ’s. Hence the order of σ is the lcm of the orders
of the cycles in its cycle decomposition.
17. Let σ ∈ Sn be (a b)(c d). There are n choices for a, n − 1 choices for b, and so on. But a, b can be
permuted, c, d can be permuted, and (a, b), (c, d) can be permuted, and so we are overcounting by a
factor of 23 = 8. Hence the number of such permutations is n(n − 1)(n − 2)(n − 3)/8.
19. Let (n) denote a cycle of length n. Then elements of S7 can be written as (7), (6), (5), (5, 2), (4),
(4, 3), (4, 2), (3), (3, 3), (3, 2, 2), (2), (2, 2, 2), corresponding to elements of order 7, 6, 5, 10, 4, 12, 3, 2 by
problem 15.
20. First note that S3 consists of the elements 1, (1 2), (1 3), (2 3), (1 2 3). Take x = (1 2 3) and note that
x3 = 1. Furthermore, for any 2-cycle y, we have y 2 = 1 and therefore y = y −1 . Then let y = (1 2),
and so we have xyx = y, hence xy = yx−1 . This is exactly the dihedral group D6 , with representation
hx, y | x3 = y 2 = 1, xy = yx−1 i, and we know |D6 | = 6. So this is a valid set of generators and
relations for S3 as well.
1.4
Matrix Groups
1. GL2 (F2 ) consists of 2 × 2 invertible matrices with entries in {0, 1}. For such a matrix to be invertible,
we must have ad − bc = 1, giving the following six possibilities. Thus |GL2 (F2 )| = 6.
1 1
1 0
1 0
0 1
1 1
0 1
,
,
,
,
,
0 1
1 1
0 1
1 1
1 0
1 0
2. Exponentiating this list of matrices, we get the corresponding list of orders 2, 2, 1, 3, 3, 2.
2
5. Suppose F has a finite number of elements. Then GLn (F ) has at most |F |n elements, since each entry
in the n × n matrix has |F | possibilities, and therefore GLn (F ) is finite. Inversely, if F has an infinite
number of elements, then to every element x ∈ F we associate the matrix xI, which has inverse x−1 I
and is therefore in GLn (F ), making GLn (F ) infinite.
4
7. Consider the two numbers in the first row of a matrix in GL2 (Fp ). If they are (0, 0), then regardless
of what the rest of the elements are, the matrix is rank-deficient and non-invertible. If they are not
(0, 0), i.e. they are one of the p2 − 1 other possibilities, there are p choices for the second row such that
the second row is a multiple of the first, making the matrix non-invertible. Hence the total number of
non-invertible matrices is p2 + p(p2 − 1) = p3 + p2 − p. Subtracting this from the total number of 2 × 2
matrices, p4 , we get |GL2 (Fp )| = p4 − p3 − p2 + p.
10. (a) The matrix multiplication works out as
a1 b1
a2 b2
a1 a2
=
0 c1
0 c2
0
a1 b2 + b1 c2
c1 c2
∈G
(b) Using the trick for 2 × 2 matrices of flipping the diagonal and negating the other diagonal,
a
0
b
c
−1
1
=
ac
−b
1/a
=
a
0
c
0
−b/(ac)
1/c
∈G
(c) Since the identity matrix I ∈ G, multiplication is closed in G, and every element has an inverse,
G is a subgroup of GL2 (R).
(d) Replacing c with a in the above proofs that multiplication is closed and every element is invertible,
we find that the elements of G with equal diagonal elements form a subgroup as well.
11. (a) Bashing out the matrix multiplication,


1
1 a b
0 1 c  0
0
0 0 1
we have
 
1
d e
1 f  = 0
0
0 1
a+d
1
0

b + e + af
c+f 
1
Hence H(F ) is closed under multiplication. Letting a = 0 and b = c = 1 for X, and d = e = 1
and f = 0 for Y , we get the top right element b + e + af = 2 6= 3 = e + b + cd, hence XY 6= Y X.
(b) In the above multiplication, if we want the right hand side to be the identity, d = −a and f = −c
and e = ac − b. Clearly these are in R, hence H(F ) closed under inverses.
(c) Using the multiplication above with a matrix Z with entries g, h, i, we get


1 a + d + g b + e + af + h + (a + d)i
 = X(Y Z)
1
c+f +i
(XY )Z = 0
0
0
1
Hence multiplication is associative, and H(F ) satisfies the axioms of a group. Since a, b, c ∈ F ,
there are |F |3 choices for them, and hence |H(F )| = |F |3 .
(d) Let’s represent the matrix with elements a, b, c as (a, b, c). Then arranging the elements in lexicographic order, starting from (0, 0, 0) and ending with (1, 1, 1), the orders are 1, 2, 2, 2, 2, 4, 2,
4.
(e) By the multiplication formula above, we can find by induction that

n 

1 a b
1 na nb + acn(n + 1)/2
0 1 c  = 0 1

nc
0 0 1
0 0
1
Hence if a or c is non-zero, then na or nc will be non-zero. On the other hand, if a = c = 0,
we must have b non-zero, and therefore nb + acn(n + 1)/2 6= 0. So every non-identity element of
H(R) has infinite order.
5
1.5
The Quaternion Group
3. hi, j, | i4 = j 4 = 1, i2 = j 2 , ji = i3 ji
1.6
Homomorphisms and Isomorphisms
1. (a) Since φ is a homomorphism, we have φ(x2 ) = φ(x)φ(x). Proceeding by induction, we get φ(xn ) =
φ(x)n for n ∈ Z+ .
(b) Note that x−1 = xk where k = |x| − 1 ∈ Z+ , otherwise x = 1 and φ(1) = φ(1) is trivially true.
Then applying part (a) to xk gives the desired result.
3. If ab = ba for a, b ∈ G, then φ(b)φ(a) = φ(ba) = φ(ab) = φ(a)φ(b), but φ is bijective, hence every
element in H can be written as φ(x) for x ∈ G. Hence H is abelian. The same argument works to
show the converse. Note that if φ is not an isomorphism, we at least need it to be surjective for this
argument to work.
7. Note that Q has only one order 2 element, −1. However, D8 contains numerous order 2 elements: r2
and sr2 , to name two. Hence D8 is not isomorphic to Q.
12. Define the map φ((a, b), c) = (a, (b, c)) = (a, b, c). This is clearly an isomorphism, by verifying that
φ(((a1 , b1 ), c1 )((a2 , b2 ), c2 )) = φ((a1 , (b1 , c1 ))(a2 , (b2 , c2 ))) through some simple algebra.
16. We have π1 ((a1 , b1 )(a2 , b2 )) = π1 ((a1 a2 , b1 b2 )) = a1 a2 = π1 ((a1 , b1 ))π1 ((a2 , b2 )), and π1 ((a, b)) = 1
when a = 1, hence ker π1 = 1 × B. The same goes for π2 , except ker π2 = A × 1.
17. If G is abelian, then φ(ab) = (ab)−1 = a−1 b−1 = φ(a)φ(b) is a homomorphism. Conversely, if φ(ab) =
φ(a)φ(b), then (ab)−1 = a−1 b−1 , and 1 = aba−1 b−1 =⇒ ba = ab and G is abelian.
20. Function composition is associative, and the composition of isomorphisms is still an isomorphism,
because each preserves group structure and is bijective. Furthermore, the identity element is the
trivial isomorphism from G to G, and the inverse of each isomorphism is simply its inverse, which
exists because an isomorphism is bijective. Hence Aut(G) is a group.
23. The map x 7→ x−1 σ(x) is injective, since if x−1 σ(x) = y −1 σ(y), then yx−1 = σ(y)σ(x−1 ) = σ(yx−1 ),
so yx−1 = 1 and y = x. But since it maps from G to G, it must be bijective. Let y = x−1 σ(x) be an
arbitrary element in G. Then σ(y −1 ) = σ(x−1 σ(x)) = σ(x−1 )x = σ(x)−1 x = (x−1 σ(x))−1 = y −1 , so σ
is actually the map σ(x) = x−1 . Since σ is an automorphism, we know from a previous exercise that
G must be abelian.
1.7
Group Actions
1. Let g1 , g2 ∈ F × , then (g1 g2 ) · a = g1 g2 a = g1 · (g2 · a) by associativity of multiplication in a field. If
g = 1, then 1 · a = 1a = a by the multiplicative identity in a field. Hence F × acts on F by g · a = ga.
4. (a) Suppose g, h ∈ G such that g ·a = a and h·a = a for all a ∈ A. Then (gh)·a = g ·(h·a) = g ·a = a,
hence the kernel is closed under the group operator. Furthermore, multiplying by g −1 in g · a = a
gives g −1 g · a = a = g −1 · a, hence the kernel is closed under inverses. Hence the kernel is a
subgroup.
(b) Take the proof above for a fixed a. Hence the stabilizer is a subgroup.
5. Suppose g · a = a for all a ∈ A. Then φg (a) = g · a = a, so φg = 1. Conversely, if φg (a) = a, then
g · a = a for all a ∈ A, so g is in the kernel of the action of G on A.
6
6. G acting faithfully on A is equivalent to the kernel of the homomorphism G → SA being trivial. But
by the previous result, that kernel is the same as the kernel of the group action, hence it is equivalent
also to the kernel of the group action being trivial.
7. Suppose α(r1 , . . . , rn ) = (r1 , . . . , rn ). Then we must have α = 1, hence the kernel of the action of F ×
on V is trivial, and the action is faithful.
8. (a) Note that (σ1 σ2 ) · {ai } = {(σ1 σ2 )(ai )} = {σ1 (σ2 (ai ))} = σ1 · (σ2 · {ai }), and 1 · {ai } = {ai }, hence
SA acting on A is indeed a group action.
(b)
•
•
•
•
•
•
(1
(1
(1
(1
(1
(1
2) · {1, 2} = {2, 1},
2) · {1, 3} = {2, 3},
2) · {1, 4} = {2, 4},
2) · {2, 3} = {1, 3},
2) · {2, 4} = {1, 4},
2) · {3, 4} = {3, 4},
and
and
and
and
and
and
(1
(1
(1
(1
(1
(1
2
2
2
2
2
2
3) · {1, 2} = {2, 3}
3) · {1, 3} = {2, 1}
3) · {1, 4} = {2, 4}
3) · {2, 3} = {3, 1}
3) · {2, 4} = {3, 4}
3) · {3, 4} = {1, 4}
11. Taking the vertices to be labeled 1, 2, 3, 4 clockwise, we have
• 1 gives 1, 2, 3, 4, which is the identity permutation 1 ∈ S4
• r gives 4, 1, 2, 3, which is the permutation (1 4 3 2) ∈ S4
• r2 gives 3, 4, 1, 2, which is the permutation (1 3)(2 4) ∈ S4
• r3 gives 2, 3, 4, 1, which is the permutation (1 2 3 4) ∈ S4
• s gives 1, 4, 3, 2, which is the permutation (2 4) ∈ S4
• sr gives 2, 1, 4, 3, which is the permutation (1 2)(3 4) ∈ S4
• sr2 gives 3, 2, 1, 4, which is the permutation (1 3) ∈ S4
• sr3 gives 4, 3, 2, 1, which is the permutation (1 4)(2 3) ∈ S4
14. Note that (g1 g2 ) · a = ag1 g2 , but g1 · (g2 · a) = ag2 g1 , and these are not necessarily equal since G is
non-Abelian. Hence this is not a group action.
16. Note that (g1 g2 ) · a = g1 g2 a(g1 g2 )−1 = g1 g2 ag2−1 g1−1 = g1 (g2 · a)g1−1 = g1 · (g2 · a), and 1 · a = a. Hence
conjugation is a group action.
19. Clearly this map H → O is onto, since if o ∈ O, then o = hx for some h ∈ H. It is also clearly injective,
since if h1 x = h2 x, we have h1 = h2 . Hence every orbit under the action of H has cardinality |H|. But
these orbits partition G, so G is some union of sets all of cardinality |H|. It follows that |G| = k|H|
for some integer k.
7
Chapter 2
Subgroups
2.1
Definition and Examples
4. Let G = (Z, +) and H = Z+ . Then H is clearly closed under +, since the sum of any two positive
integers is still a positive integer, but H is not closed under inverses, so H is not a subgroup of G.
5. Let g ∈ G but g 6∈ H, and 1 6= h ∈ H (we find such an h since |H| > 1). Then x = gh ∈ H since
gh 6= g. Since H is a subgroup, it is closed under inverses, so h−1 ∈ H. Then g = xh−1 ∈ H, and by
contradiction, |H| =
6 n − 1. Alternatively, we can use Lagrange’s theorem, since n − 1 - n.
6. Note that |1| = 1 < ∞, and if g n = 1, then (g −1 )n = (g n )−1 = 1 (here we use the abelian-ness of G),
so |g −1 | < ∞. Let hm = 1, then (gh)lcm(n,m) = 1 by a previous result, so |gh| < ∞. Thus the torsion
subgroup is a subgroup if G is abelian. To see that the torsion subgroup is not actually a subgroup
when G is not abelian, take G = GL2 (R) and observe that
−1 1
−1 0
1 1
=
0 1
0 1
0 1
which has infinite order, but the terms of the product both have order 2.
7. Every element except 0 in Z has infinite order, but every element of Z/nZ is of finite order, since
there are finitely many elements in Z/nZ. Hence the torsion subgroup is 0 × Z/nZ. However, (1, 0̄)
and (−1, 0̄) both have infinite order, but their sum (0, 0̄) has finite order, so the set of infinite-order
elements with the identity is not a subgroup.
8. Without loss of generality, let |K| >= |H|. If H ⊆ K, then H ∪ K = K, which is a subgroup. Now
suppose there exists h ∈ H 6∈ K and k ∈ K 6∈ H (otherwise H ⊆ K). Then hk 6∈ H, otherwise we write
x = hk ∈ H and k = xh−1 ∈ H, and similarly, hk 6∈ K, otherwise x = hk ∈ K and h = xk −1 ∈ K.
Hence H ∪ K is not closed, and therefore not a subgroup.
13. Suppose H 6= 0, and note that if x ∈ H, then nx ∈ H where n ∈ Z, since nx = x + . . . + x for positive
n, and nx = (−x) + . . . + (−x) for negative n. Let a/b ∈ H. Then b(a/b) = a ∈ H and a(b/a) = b ∈ H.
By Bezout’s lemma, there exists integers x, y ∈ Z such that ax+by = 1, hence 1 ∈ H. Then n, 1/n ∈ H
and m/n ∈ H for m, n ∈ Z. Thus H = Q.
14. Note that srsr = ssr−1 r = 1 and sr2 sr2 = srsr−1 r2 = srsr = 1, but srsr2 = ssr−1 r2 = r, which has
order n ≥ 3, so the given set is not a subgroup of D2n .
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