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Functions and Groups
The Fundamentals of Functions
A function $f: A \to B$ is a rule assigning each element in a Domain to a unique element in a Codomain.
$$f: A \to B$$
Describes the mapping of elements from the source set to the target set.
$A$=The Domain; set of all possible inputs(set)
$B$=The Codomain; set of all potential outputs(set)
$f(A)$=The Range; the actual set of outputs reached(subset of B)
Domain: The collection of all valid input values (pre-images).
Codomain: The entire set $B$ into which the function is defined to map.
Range: The Image of the entire domain, representing the actual outputs used in $B$.
Classification of Mappings
Mappings are categorized as Injective (one-to-one), Surjective (onto), or Bijective based on their element pairings.
$$\forall a_1, a_2 \in A, f(a_1) = f(a_2) \implies a_1 = a_2$$
The formal definition of an injective (one-to-one) function.
$a_1, a_2$=Distinct elements of the domain(elements)
$f(a)$=The image of element 'a' under mapping 'f'(output)
Injective: Each input maps to a unique, distinct output. Passed by the Horizontal Line Test.
Surjective: Every element in the codomain is mapped to by at least one element in the domain.
Bijective: A one-to-one correspondence where the function is both injective and surjective.
Binary Operations and Group Axioms
A set $G$ with a Binary Operation $\star$ is a group if it satisfies four essential axioms.
$$(G, *) \text{ is a group if } G_1, G_2, G_3, G_4$$
The structural conditions for a set to be an algebraic group.
$G_1$=Closure: result stays in the set(axiom)
$G_2$=Associativity: $(a * b) * c = a * (b * c)$(axiom)
$G_3$=Identity: Existence of element 'e'(axiom)
$G_4$=Inverse: Every element has a partner to reach 'e'(axiom)
Closure: For all $a, b \in G, a \star b \in G$.
Associativity: For all $a, b, c \in G$, $(a \star b) \star c = a \star (b \star c)$.
Identity: There exists $e \in G$ such that $a \star e = e \star a = a$.
Inverse: For every $a \in G$, there is $a^{-1} \in G$ such that $a \star a^{-1} = e$.
Reversal Law: $(a \star b)^{-1} = b^{-1} \star a^{-1}$ — the inverse of a product reverses the order.
Abelian Groups and Group Order
An Abelian group satisfies the Commutative law, meaning the order of operation does not change the result.
$$a * b = b * a$$
Commutativity: The defining feature of Abelian groups. Visualized as symmetry across the main diagonal of a Cayley Table.
Order of a Group: Represented as $|G|$, this is the total count of elements in the set.
Finite vs Infinite: A Finite Group has a specific integer order, while an Infinite Group (like integers under addition) does not.
Composition and Inverse of Functions
The Composition of two functions $f$ and $g$ is written as $(g \circ f)(x) = g(f(x))$ — apply $f$ first, then feed the result into $g$.
$(g \circ f)(x) = g(f(x))$
The output of f becomes the input of g.
$f$=Inner function — applied first
$g$=Outer function — applied to the result of f
Order matters: $g \circ f$ means apply $f$ first, then $g$. Do not confuse with $f \circ g$.
Verification: $(f^{-1} \circ f)(x) = x$ and $(f \circ f^{-1})(x) = x$ — composing a function with its inverse gives the identity.
Shorthand: $f^2(x) = f(f(x))$, $f^3(x) = f(f(f(x)))$, and so on.
An Inverse Function $f^{-1}$ reverses the mapping of $f$: if $f(a) = b$, then $f^{-1}(b) = a$. Only Bijective functions have inverses.
$f^{-1}(f(x)) = x$
Applying f then f⁻¹ (or vice versa) returns the original input.
$f^{-1}$=The inverse function of f
$f$ is not bijective
→$f^{-1}$ does not exist as a function
$f$ is a linear function $y = mx + c$
→The inverse is always another linear function
Algebraic method: Set $y = f(x)$, swap $x$ and $y$, solve for $y$ — this gives $f^{-1}(x)$.
Domain-Range swap: Domain of $f^{-1}$ = Range of $f$, and Range of $f^{-1}$ = Domain of $f$.
Graphical symmetry: The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$.
Algebraic Structure Hierarchy
Algebraic structures build up in layers: a Groupoid gains associativity to become a Semi-group, adds an identity to become a Monoid, and adds inverses to become a Group.
Groupoid: A set closed under a binary operation (closure only).
Semi-group: A groupoid where the operation is also associative.
Monoid: A semi-group that has an identity element $e$.
Group: A monoid where every element has an inverse.
Key non-examples: $(\mathbb{N}, +)$ is a semi-group (no identity 0), $(\mathbb{Z}, \times)$ is a monoid (identity 1, but most elements lack integer inverses).
Application Strategy
Worked Examples
Problem Statement
Determine the inverse function $f^{-1}(x)$ for the bijective function $f(x) = \frac{3x - 5}{2}$.
Inverse functionsAlgebraic manipulation
Strategic Toolkit
$y = f(x) \leftrightarrow x = f^{-1}(y)$
Step 1
Set notation
Set the function equal to $y$.
$y = \frac{3x - 5}{2}$
Step 2
Swap variables
Swap the variables $x$ and $y$ to begin finding the inverse.
$x = \frac{3y - 5}{2}$
Step 3
Isolate numerator
Multiply both sides by 2.
$2x = 3y - 5$
Step 4
Isolate term
Add 5 to both sides.
$2x + 5 = 3y$
Step 5
Solve for y
Divide by 3 to isolate $y$.
$y = \frac{2x + 5}{3}$
Step 6
Final Form
State the final inverse function.
$f^{-1}(x) = \frac{2x + 5}{3}$
Problem Statement
Verify if the set $S = \{1, -1\}$ under standard multiplication forms a group.
Group AxiomsClosureIdentityInverse
Strategic Toolkit
$(G, *) \text{ Axioms}$
Step 1
Check Closure
Multiply all pairs: $1 \times 1 = 1$, $1 \times -1 = -1$, $-1 \times -1 = 1$.
$1, -1 \in S \text{ (Check)}$
Step 2
Check Associativity
Multiplication of real numbers is always associative.
$a(bc) = (ab)c \text{ (Check)}$
Step 3
Identify Identity
$1 \times x = x$ for all $x$ in $S$.
$e = 1 \in S \text{ (Check)}$
Step 4
Identify Inverses
Find elements that multiply to 1: $1 \times 1 = 1$ and $-1 \times -1 = 1$.
$1^{-1}=1, (-1)^{-1}=-1 \text{ (Check)}$
Step 5
Conclusion
Since all four axioms are satisfied, $(S, \times)$ is a group.
$\text{Group Verified}$
Problem Statement
Let $f: \{1,2,3\} \to \{a,b,c\}$ be defined by $f = \{(1,a),(2,b),(3,c)\}$. Classify this function.
InjectiveSurjectiveBijective
Strategic Toolkit
$f(a_1)=f(a_2) \implies a_1=a_2 \text{ (injective)}, \text{Range}=\text{Codomain (surjective)}$
Step 1
Check Injective
Are all outputs distinct? $f(1)=a, f(2)=b, f(3)=c$ — no two inputs share an output.
$a \neq b \neq c$ — Injective ✓
Step 2
Check Surjective
Is every element of the codomain $\{a,b,c\}$ hit? Range $= \{a,b,c\}$ = Codomain.
$\text{Range} = \{a,b,c\} = \text{Codomain}$ — Surjective ✓
Step 3
Conclusion
Since $f$ is both injective and surjective, it is bijective — a perfect one-to-one correspondence.
$f$ is bijective
Memorize formulas
Identity Property
The element 'e' that leaves other elements unchanged during the operation.
Formula
a * e = e * a = a
Inverse Property
The element 'a⁻¹' that returns the system to the identity.
Formula
$a * a^{-1} = a^{-1} * a = e$
Composite Function
Applying function g to the result of function f.
Formula
$(g \circ f)(x) = g(f(x))$
Abelian Condition
Used when the order of the operation does not matter (Commutativity).
Formula
a * b = b * a
Reversal Law of Inverses
The inverse of a product reverses the order of the factors.
Formula
$(a * b)^{-1} = b^{-1} * a^{-1}$
Review
1
A Cayley table for a group $(G, \star)$ is symmetric about its main diagonal. This tells you that the group is:
Ans
Abelian — because symmetry means $a \star b = b \star a$ for all elements
Review
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Read symmetry as commutativity
A symmetric Cayley table means the entry in position $(a,b)$ matches the entry in position $(b,a)$ for every pair of group elements. That is exactly the statement that the operation does not depend on order.
$$a \star b = b \star a \quad \text{for all } a,b \in G$$
2
Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = |x|$. Which of the following is true about this function?
Ans
It is neither injective nor surjective
Review
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Check injectivity first
Two different inputs can give the same output under absolute value, so the function fails the one-to-one test.
$$f(3) = |3| = 3 \quad \text{and} \quad f(-3) = |-3| = 3$$
Check the range next
Absolute value never produces a negative number, so values like $-1$ cannot be reached. That means the function is not onto $\mathbb{R}$ either.
$$f(x) = |x| \ge 0 \quad \Rightarrow \quad -1 \notin \text{Range}(f)$$
3
In the group $(\mathbb{Z}, +)$, what is the identity element and what is the inverse of an element 'a'?
Ans
Identity is 0, Inverse is -a
Review
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Use the additive identity
In $(\mathbb{Z}, +)$, the number that leaves every integer unchanged under addition is $0$.
$$a + 0 = a$$
Solve for the inverse
The inverse of $a$ is the number that adds with it to give the identity. For addition, that number is $-a$.
$$a + (-a) = 0$$
4
The set $\{1, 2, 3, 4, 5\}$ under addition does NOT form a group. Which group axiom fails first?
Ans
Closure — because $3 + 4 = 7$, which is not in the set
Review
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Test closure with a simple pair
A set under addition must contain the sum of any two of its elements. Here, adding two members already leaves the set.
$$3 + 4 = 7 \notin \{1,2,3,4,5\}$$
5
If $f(x) = 2x + 1$ and $g(x) = x^2$, what is the value of the composite function $(f \circ g)(3)$?
Ans
19
Review
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Apply the inner function first
In a composite, evaluate the inside function before the outside one. So compute $g(3)$ first.
$$g(3) = 3^2 = 9$$
Feed that value into $f$
Now substitute the result into $f(x)=2x+1$.
$$(f \circ g)(3) = f(9) = 2(9) + 1 = 19$$
6
If $f(x) = x + 3$ and $g(x) = 2x$, what is $(g \circ f)(x) - (f \circ g)(x)$?
Ans
3
Review
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Compose in the first order
Substitute $f(x)=x+3$ into $g(x)=2x$.
$$(g \circ f)(x) = g(x+3) = 2(x+3) = 2x+6$$
Compose in the reverse order and subtract
Now substitute $g(x)=2x$ into $f(x)=x+3$, then take the difference of the two expressions.
$$(f \circ g)(x) = f(2x) = 2x+3, \qquad (g \circ f)(x) - (f \circ g)(x) = (2x+6) - (2x+3) = 3$$
7
If $f(x) = 2x + 6$ and $g(x) = \frac{x - 6}{2}$, what is $f(g(x))$?
Ans
$x$ — confirming that $g = f^{-1}$
Review
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Substitute the inner function
Replace the input of $f$ with $g(x)=\frac{x-6}{2}$ and simplify the result step by step.
$$f(g(x)) = 2\left(\frac{x-6}{2}\right) + 6 = x - 6 + 6$$
Finish the cancellation
The constants cancel, leaving the identity function. That is exactly what must happen for an inverse composition.
$$f(g(x)) = x$$
8
If $f(x) = \sqrt{x - 2}$ has domain $[2, \infty)$ and range $[0, \infty)$, what are the domain and range of $f^{-1}$?
Ans
Domain $[0, \infty)$, Range $[2, \infty)$
Review
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Swap domain and range
For any inverse function, the original range becomes the new domain, and the original domain becomes the new range.
$$\operatorname{dom}(f^{-1}) = [0, \infty), \qquad \operatorname{range}(f^{-1}) = [2, \infty)$$
9
A function $f: A \to B$ where $|A| = 5$ and $|B| = 3$ can be:
Ans
Surjective but not injective
Review
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Use the pigeonhole principle
Because the domain has more elements than the codomain, at least two inputs must land on the same output. So injectivity is impossible.
$$|A| = 5 > 3 = |B| \Rightarrow f \text{ cannot be injective}$$
Check surjectivity separately
A function can still hit every element of the smaller codomain, so onto is possible even though one-to-one is not.
$$f \text{ can be surjective but not injective}$$
10
In a group, if $a^{-1} = 3$ and $b^{-1} = 5$, then $(a \star b)^{-1}$ equals:
Ans
$b^{-1} \star a^{-1} = 5 \star 3$
Review
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Apply the reversal law
The inverse of a product in a group is the product of the inverses in reverse order.
$$(a \star b)^{-1} = b^{-1} \star a^{-1}$$
Substitute the given inverses
Replace $a^{-1}$ with $3$ and $b^{-1}$ with $5$.
$$(a \star b)^{-1} = 5 \star 3$$
11
In the group $(\{1, 2, 3, 4, 5, 6\}, \times_7)$ (multiplication modulo 7), what is the inverse of 3?
Ans
5
Review
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Set up the congruence
An inverse modulo 7 must multiply with 3 to give 1 modulo 7.
$$3x \equiv 1 \pmod{7}$$
Test the candidate
Checking $x=5$ gives a product that differs from 1 by a multiple of 7, so it works.
$$3 \cdot 5 = 15 \equiv 1 \pmod{7}$$
12
The set of natural numbers $\mathbb{N}$ under multiplication has closure, associativity, and identity (1), but not every element has an inverse. This structure is called a:
Ans
Monoid
Review
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Match the axioms to the structure name
A semigroup has closure and associativity. Once an identity is also present, the structure is a monoid.
$$\text{closure} + \text{associativity} + \text{identity} \Rightarrow \text{monoid}$$
Check why it is not a group
Groups also require inverses for every element, but most natural numbers do not have multiplicative inverses inside $\mathbb{N}$.
$$2^{-1} = \frac{1}{2} \notin \mathbb{N}$$
13
Which type of function is guaranteed to have an inverse that is also a function?
Ans
Bijective
Review
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Require both one-to-one and onto
An inverse must assign exactly one input to each output and must cover the whole codomain. That only happens when the original function is bijective.
$$\text{inverse is a function} \iff \text{bijective}$$
14
The set of all $2 \times 2$ non-singular real matrices under multiplication forms a group. Why is this group NOT Abelian?
Ans
Matrix multiplication is not commutative: $AB \neq BA$ in general
Review
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Identify the failing property
This set still satisfies the group axioms, so the only way it can fail to be Abelian is if the operation is not commutative.
$$AB \neq BA \quad \text{in general}$$
Conclude non-Abelian status
Since multiplication of matrices depends on order, the group is not Abelian even though it is a valid group.
$$\text{non-commutative} \Rightarrow \text{non-Abelian}$$
Question 1 of 14
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1
A Cayley table for a group $(G, \star)$ is symmetric about its main diagonal. This tells you that the group is: