SOLUTIONS 2 - MAT220
Problem 1. (a) Let G be a cyclic group and φ : G → H a group homomorphism. Show that φ is completely
determined by its value on a generator.
(b) How many group homomorphisms Z → Z2 are there?
(c) Let φ : Z8 → Z32 be a group homomorphism such that φ(1) = 4. Compute φ(5). What is the kernel of
φ?
(d) Is there a nontrivial group homomorphism Z7 → Z5 ? What about Z2 → Z? If yes, give an example. If
no, prove it. (Hint: Use exercise (a). Where does the identity go?)
Solution. (a) Let g ∈ G be a generator. Then any element x of G is of the form g n , so that
φ(x) = φ(g n ) = φ(g)n
where the latter equality follows from the homomorphism property. Hence φ is determined by its value on
a generator.
(b) By (a), we need only consider the value on a generator, e.g. 1. We may map 1 to 0 or 1 in Z2 ,
the first being the trivial homomorphism, the second sending all even integers to 0 and all odd integers to
1. Both are well-defined homomorphisms as is readily checked.
(c) Since φ(1) = 4, φ(5) = φ(1 + 1 + 1 + 1 + 1) = 5φ(1) = 20. The kernel of φ is {0}.
(d) There is no nontrivial group homomorphism Z7 → Z5 . For the sake of contradiction, suppose there
is. Then the generator 1 must be sent to a non-zero element g in Z5 . We then have φ(1) = g 6= 0, and so
φ(7) = 7g = 0. Then 7g = 2g = 0, but g is a non-zero element of Z5 , hence a generator. Then g is of order
5, contradicting 2g = 0.
There is no nontrivial group homomorphism Z2 → Z either. Suppose the generator 1 is mapped to a
nonzero element of Z, say φ(1) = n 6= 0. Then φ(2) = φ(0) = 2n 6= 0, a contradiction.
Problem 2. Classify the following groups according to the fundamental theorem of finitely generated abelian
groups.
(a) (Z × Z × Z2 )/h(3, 0, 0)i
(b) (Z × Z × Z)/h(7, 7, 7)i
(c) (Z × Z × Z)/h(n, n, n)i where n is a nonzero integer.
(d) (Z3 × Z2 )/h(2, 1)i
(e) (Z9 × Z10 )/h(3, 2)i
Solution. (a) There is a group homomorphism
φ : Z × Z × Z2 → Z3 × Z × Z2
by the identity in the last two factors and the projection Z → Z3 in the last. This is surjective (onto) and
has kernel generated by (3, 0, 0). Hence the factor group is isomorphic to Z3 × Z × Z2 by Theorem 14.11.
(b) There is a group homomorphism
φ : Z × Z × Z → Z7 × Z × Z
sending (n1 , n2 , n3 ) to ([n1 ], n2 − n1 , n3 − n2 ) where [n1 ] is the equivalence class of n1 in Z7 . This is a
surjection: for (m1 , m2 , m3 ) in Z7 × Z × Z, it is hit by (7k1 + m1 , 7k1 + m1 + m2 , 7k1 + m1 + m2 + m3 ),
where k1 is any integer. The kernel is then generated by (7, 7, 7), as the first factor must be a multiple of 7,
and the last two factors demand n2 = n1 and n3 = n2 for an element of the kernel. Hence the factor group
is isomorphic to Z7 × Z × Z by Theorem 14.11.
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SOLUTIONS 2 - MAT220
(c) By the same argument as above, the factor group is isomorphic to Zn × Z × Z.
(d) 2 is of order 3 in Z3 and 1 is of order 2 in Z2 , so (2, 1) is of order 6 in Z3 × Z2 . Hence the factor
group is trivial.
(e) 3 is of order 3 in Z9 , 2 is of order 5 in Z10 , so (3, 2) is of order 15 in Z9 × Z10 . Hence the factor
group is of order six, and there is only one such abelian group up to isomorphism.
Problem 3. In Exercise 9, Section 16, the definition of isomorphic G-sets is given. Define a G-map
be a map between G-sets φ : X → Y satisfying φ(gx) = gφ(x), however we do not demand the map to
surjective or injective. Let X = {1, 2, 3, 4} and Y = {1, 2, 3, 4, 5}. Let S4 act on X by σ.x = σ(x), while
acts on Y by σ.y = σ(y) if y 6= 5, while σ.5 = 5 for all σ ∈ S4 .
(a) Show that the proposed action of S4 on Y is a group action.
(b) Define φ : X → Y by

1 7→ 1



2 7→ 5

3 7→ 4



4 7→ 2
Determine if φ is an S4 -map.
(c) Set X = Y = {1, . . . , n}. Let Sn act on X by σ.x = σ(x). Let τ ∈ Sn be arbitrary. Is τ always
Sn -map?
(d) In the setting above, is τ always a hτ i-map?
Solution. (a) The identity fixes everything. For associativity:
(
(
σ.τ (y) if y 6= 5
σ(τ (y)) if y 6= 5
σ.(τ.y) =
=
σ.5 if y = 5
5 if y = 5
to
be
S4
an
= (στ ).y.
(b) This is not an S4 -map. Let σ = (2, 3). Then φ(σ.2) = φ(σ(2)) = φ(3) = 4, while σ.φ(2) = σ.5 = 5.
(c) τ , considered as a map τ : X → X, is not always an Sn -map, as Sn is not commutative for n ≥ 3.
(d) τ is a hτ i-map, as τ commutes with τ n for all integers n.
Problem 4. On a regular octagon, the vertices are labeled with integers between 1 and 8, where one number
may not be used more than once.
(a) How many ways are there to number vertices in total?
(b) Two ways of labeling the vertices are equivalent if one is obtained from the other by a rotation of the
octagon. How many non-equivalent ways are there of labeling the points?
(c) Answer both (a) and (b), however allow for the same number to be used repeatedly, i.e. two or more of
the vertices may be assigned the same number between 1 and 8.
Solution. (a) 8!.
(b) We let X be the set of different labelings of the octagon. Then |Xe | = 8! (here we act by rotations
with Z8 ). As the same number may not be used more than once, we have that |Xg | = 0 for all non-identity
elements. Hence, by Burnside, we have 7! non-equivalent ways of labeling the points.
(c) If we allow for the same number to be used repeatedly, there are 88 ways to label the vertices in
total. Note that |Xn | = 8gcd(n,8) with n ∈ Z8 . Hence there are
7
1 8 X gcd(i,8)
(8 +
8
)
8
i=1
different ways of labeling the vertices.
SOLUTIONS 2 - MAT220
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Problem 5. A sequence of groups (Gi )i∈Z and morphisms {φi : Gi → Gi−1 }, i.e. a sequence
···
φi+2
Gi+1
φi+1
Gi
φi
Gi−1
φi−1
···
is called exact if im φi+1 = ker φi for all i ∈ Z. Define a group homomorphism f : Z → Z by f (n) = 2n. Fill
in the blanks such that the sequence
···
0
0
f
Z
Z
−
−
0
0
···
is exact. (Hint: What is the image of f ? The zero after the blank forces your chosen homomorphism to
have a certain property.).
Solution. Call our chosen function g and our chosen group G. The kernel of G → 0 is all of G, and so the
image of g must be all of G, i.e. g must be surjective. The image of f consists of all even integers, and so
the kernel of g must consist of all even integers. We see that g : Z → Z2 works.
Problem 6. Let G be a group, H a subgroup of G and N a normal subgroup of G.
(a) Prove that the product HN = {hn | h ∈ H, n ∈ N } is a subgroup of G.
(b) Prove that H ∩ N is a normal subgroup of H.
(c) Prove that there is an isomorphism HN/N ∼
= H/(N ∩ H). (Hint: Use Theorem 14.11.)
Solution. (a) Clearly the identity is in HN . To see that it is closed, consider a product (hn)(h0 n0 ). We
rewrite this as
(hn)(h0 n0 ) = hh0 h0−1 nh0 n0 = hh0 ñn0
where h0−1 nh0 = ñ ∈ N since N is normal. For inverses, note that
(hn)−1 = n−1 h−1 = h−1 hn−1 h−1 = h−1 n0
for some n0 ∈ N.
(b) We already know that the intersection of two subgroups is a subgroup. To see that H ∩ N is normal in H, let h1 , h2 ∈ H and n ∈ H ∩ N . Then h1 nh−1
is clearly in H, as all three elements are in H.
2
Moreover, it is necessarily in N as N is normal in G.
(c) Restrict φ : G → G/N to H and HN respectively, and consider them as functions onto their images.
Then φH : H → φ[H] and φHN : HN → φ[HN ] = φ[H] have the same image, and the kernels are H ∩ N
and N , respectively. This concludes the argument by Theorem 14.11.
Problem 7. Let G be a group with two given subgroups, H1 and H2 . Define the following equivalence
relation ∼ on G: g1 ∼ g2 if and only if g1 = h1 g2 h2 for some h1 ∈ H1 , h2 ∈ H2 .
(a) Prove that the proposed equivalence relation is an equivalence relation. An equivalence class under this
relation is called a double coset, and the collection of all double cosets is denoted H1 \G/H2 . The equivalence
class of g is denoted H1 gH2 .
(b) Define a function
(H1 × H2 ) × G → G
by ((h1 , h2 ), g) 7→ h1 gh−1
2 Prove that this is a group action of the group H1 × H2 on the set G.
(c) (Note: You are not asked to compute anything in this exercise. Keep all answers short.) Assume all
groups involved are finite. Why does Burnside’s formula give the number of double cosets? Suppose you
were to compute the number of double cosets using Burnside’s formula, which in general looks like this:
1 X
|Xk |.
|K|
k∈K
In this case, what is K and Xk ?
Solution. (a) We have that g1 ∼ g1 , since we may pick h1 = h2 = e. Now, if g1 ∼ g2 , then g1 = h1 g2 h2 , so
−1
g2 = h−1
1 g1 h2 . Hence g2 ∼ g1 . Lastly, suppose g1 ∼ g2 and g2 ∼ g3 , so that g1 = h1 g2 h2 and g2 = h3 g3 h4 .
Then g1 = h1 (h3 g3 h4 )h2 = (h1 h3 )g3 (h4 h2 ). We conclude that ∼ is an equivalence relation.
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SOLUTIONS 2 - MAT220
(b) The identity fixes everything. For associativity, write (h1 , h2 ).g = h1 gh−1
2 . Then
−1 −1
−1
(h3 , h4 ).((h1 , h2 ).g) = (h3 , h4 ).(h1 gh−1
= (h3 h1 , h4 h2 ).g.
2 ) = h3 h1 gh2 h4 = h3 h1 g(h4 h2 )
We conclude that this is a group action.
(c) Burnside’s formula gives us the number of double cosets as the number of double cosets are precisely the
number of orbits under this equivalence relation. In this case, K = H1 × H2 , and |Xk | are those elements of
G fixed by the action of k ∈ H1 × H2 .