Mathematics 1214: Introduction to Group Theory
Solutions to tutorial exercise sheet 1
1. Let m, n ∈ N, and let S = {1, 2, . . . , m} and T = {1, 2, . . . , n}. How many
mappings are there from S to T ? How many of these are injective, and how many
are bijective?
Solution Consider the problem of choosing the values of an arbitrary mapping
α : S → T . There are n possible values of α(1), n possible values of α(2),. . . , and
n possible choices of α(m). The total number of possible choices is therefore
m
n
{z· · · × n} = n .
| ×n×
m times
So there are nm mappings S → T .
If α : S → T is injective, then {α(1), α(2), . . . , α(n)} are distinct elements of T . So
T contains at least n elements, so n ≥ m. In other words, if n < m then there are
no injective mappings S → T .
Now suppose that n ≥ m and consider the problem of choosing the values of an
arbitrary injective mapping α : S → T . There are n possible values of α(1), and
since α is injective α(2) must be chosen from the n − 1 elements of T excluding
α(1). Similarly, α(3) must be chosen from the n − 2 elements of T excluding both
α(1) and α(2). Continuing this argument, we see that the total number of possible
choices is
n!
.
n × (n − 1) × (n − 2) × . . . = n × (n − 1) × · · · × (n − m + 1) =
|
{z
}
(n − m)!
m terms
So if n ≥ m then there are n!/(n − m)! injective mappings S → T .
If m < n then {α(1), α(2), . . . , α(m)} is a set of size at most m, so cannot be all
of T . So there is no surjective mapping S → T if m < n, and we’ve seen above
that there’s no injective mapping S → T if n < m. So if m 6= n then there are no
bijections S → T .
If n = m and α : S → T is an injective map, then {α(1), . . . , α(m)} are m = n
distinct elements of T , so {α(1), . . . , α(m)} = T and α is surjective. Hence every
injection is a bijection in this case. So the number of bijections S → T is the same
as the number of injections S → T , which is n!/(n − n)! = n!.
2. Let m ∈ N. How many surjective mappings are there from {1, 2, . . . , m} to
{1, 2, 3}? [Hint: how many mappings are there? How many of these are not
surjective?]
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Solution Let S = {1, 2, . . . , m}. There are 3m mappings S → {1, 2, 3}. For
k = 1, 2, 3, Let Mk be the set of mappings α : S → {1, 2, 3} for which α(x) 6= k for
every x ∈ S. Then the non-surjective mappings are M1 ∪ M2 ∪ M3 . Observe that
|M1 ∪M2 ∪M3 | = |M1 |+|M2 |+|M3 |− |M1 ∩M2 |+|M2 ∩M3 |+|M3 ∩M1 | +|M1 ∩M2 ∩M3 |.
[To understand this formula, draw a Venn diagram and think about how to count
the size of three sets which intersect one another]. Clearly, |M3 | is the same as the
number of mappings S → {1, 2}, which is 2m . By symmetry, |M1 | = |M2 | = |M3 |.
If α ∈ M2 ∩ M3 then α(x) 6= 2 and α(x) 6= 3 for all x ∈ S, so α(x) = 1 for
all α ∈ S. So the only function in M2 ∩ M3 is the constant function taking
the value 1, hence |M2 ∩ M3 | = 1. Similarly, |M1 ∩ M2 | = |M3 ∩ M1 | = 1. If
α ∈ M1 ∩ M2 ∩ M3 then α : S → {1, 2, 3} but α(x) 6= 1, α(x) 6= 2 and α(x) 6= 3 for
all x ∈ S, which is impossible. So M1 ∩M2 ∩M3 = ∅. Putting this all together gives
|M1 ∪ M2 ∪ M3 | = 3 · 2m − 3, so the number of surjective mappings S → {1, 2, 3}
is 3m − 3 · 2m + 3.
3. Let S, T be sets and let α : S → T . Prove that α is onto if and only if there is a
function β : T → S such that α ◦ β = ιT .
Solution First, suppose that α is onto. If y ∈ T , then since α is onto, there is
at least one x ∈ S with α(x) = y. We write β(y) for one of these x values.
We claim that β is then a mapping T → S and α ◦ β = ιT . Indeed, for every y ∈ T
we have given precisely one value x ∈ S associated by β to y, so β is a mapping,
and α ◦ β : T → T, y 7→ α(β(y)) = y by the definition of β. So α ◦ β = ιT .
Conversely, suppose that there is a function β : T → S such that α ◦ β = ιT . If
y ∈ T then y = ιT (y) = α(β(y)) and β(y) ∈ S, so y = α(x) for some x ∈ S
(namely, x = β(y)). Hence α is onto.
4. Let S be a set, and let Sym(S) be the set of all bijections α : S → S.
Prove that Sym(S) is closed under composition.
[This means: prove that if α, β ∈ Sym(S), then α ◦ β ∈ Sym(S).]
Solution Let α, β ∈ Sym(S). Since α is onto, if y ∈ S then y = α(y 0 ) for some
y 0 ∈ S. Since β is onto, y 0 = β(x) for some x ∈ S. Hence y = α(β(x)) = (α ◦ β)(x),
which shows that α ◦ β is onto.
If α ◦ β(x1 ) = α ◦ β(x2 ) for some x1 , x2 ∈ S, then α(β(x1 )) = α(β(x2 )). Since α
is one-to-one, this implies that β(x1 ) = β(x2 ). Since β is one-to-one, this implies
that x1 = x2 . Thus α ◦ β is one-to-one.
Hence α ◦ β is a bijection, so is in Sym(S).
5. Let S, T be sets. If α : S → T , let
Γ(α) = {(x, y) ∈ S × T : x ∈ S, y = α(x)}.
Γ(α) is called the graph of α.
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(a) Prove that if α and β are mappings S → T , then α = β if and only if
Γ(α) = Γ(β).
(b) Let G ⊆ S × T . Prove that G is the graph of some mapping α : S → T if
and only if for every x ∈ S, there is exactly one element y ∈ T such that
(x, y) ∈ G.
Solution (a) If α = β then α(x) = β(x) for all x ∈ S, so
Γ(β) = {(x, y) ∈ S×T : x ∈ S, y = β(x)} = {(x, y) ∈ S×T : x ∈ S, y = α(x)} = Γ(α).
Conversely, suppose that α : S → T and β : S → T are mappings with Γ(α) = Γ(β).
For every x ∈ S, we have that (x, α(x)) is in Γ(β), so by the definition of Γ(β),
this implies that α(x) = β(x). Hence α = β.
(b) Suppose that G = Γ(α) for some mapping α : S → T . For every x ∈ S, there
is exactly one y ∈ T such that y = α(x). Hence (by the definition of Γ(α)), for
every x ∈ S, there is exactly one y ∈ T such that (x, y) ∈ Γ(α) = G.
Conversely, suppose that G is a subset of S × T such that for every x ∈ S, there
is exactly one element y ∈ T with (x, y) ∈ G. Given x ∈ S, let α(x) the element
of T such that (x, y) ∈ G. In view of the property we are assuming G satisfies, α
is then a mapping S → T and we have G = Γ(α).
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