Abstract Algebra (AQA A-Level Further Maths): Revision Notes

Worked Example: Checking for Binary Operations

Question: Which of these functions are binary operations on the set ℤ?

a) $a \circledast b = a - b$

b) $a \blacksquare b = \frac{a}{b}$

Solution:

Part a: For subtraction on ℤ

Any two integers can be subtracted to give another integer. For example:

• $5 - 3 = 2$ (which is in ℤ)
• $-7 - 4 = -11$ (which is in ℤ)

Since the result always stays within ℤ, subtraction is a binary operation on ℤ.

Part b: For division on ℤ

Division does not, in general, produce a member of ℤ. Consider the example:

• $a = 2$ and $b = 3$
• $a \blacksquare b = \frac{2}{3}$

Since $\frac{2}{3}$ is not an integer, it's not a member of ℤ. Additionally, division by zero is undefined, which further prevents this from being a binary operation.

Therefore, $\blacksquare$ is not a binary operation on ℤ.

Exam tip: When checking if something is a binary operation, you only need one counter-example to prove it isn't. However, to prove it is a binary operation, you must show it works for all possible pairs of elements.

Worked Example: Testing Commutativity

Question: Which of these binary operations on the set ℕ are commutative?

a) $a \circledast b = a^b$

b) $a \circ b = a + b$

Solution:

Part a: Testing $a \circledast b = a^b$

To test commutativity, we need to check if $a^b = b^a$ for all values of $a$ and $b$ in ℕ.

Let's try a counter-example: $a = 2$ and $b = 5$

• $a \circledast b = 2^5 = 32$
• $b \circledast a = 5^2 = 25$

Since $32 \neq 25$, we have $a^b \neq b^a$ in general.

Therefore, $\circledast$ is not commutative.

Part b: Testing $a \circ b = a + b$

Addition of natural numbers is commutative because the sum doesn't depend on the order. For any $a, b \in \mathbb{N}$:

$a \circ b = a + b = b + a = b \circ a$

Therefore, $\circ$ is commutative.

Exam tip: To prove an operation is NOT commutative, give a single counter-example. To prove it IS commutative, you must show it works for all possible pairs (often using algebraic reasoning).

Worked Example: Testing Associativity

Question: Which of these binary operations on ℝ is associative?

a) $a \triangleright b = 2a + b$

b) $a \circledast b = a \times b$

Solution:

Part a: Testing $a \triangleright b = 2a + b$

We need to check if $(a \triangleright b) \triangleright c = a \triangleright (b \triangleright c)$.

Left side: $(a \triangleright b) \triangleright c$

• First, calculate $a \triangleright b = 2a + b$
• Then, $(2a + b) \triangleright c = 2(2a + b) + c = 4a + 2b + c$

Right side: $a \triangleright (b \triangleright c)$

• First, calculate $b \triangleright c = 2b + c$
• Then, $a \triangleright (2b + c) = 2a + (2b + c) = 2a + 2b + c$

Comparing: $4a + 2b + c \neq 2a + 2b + c$ in general.

Therefore, $\triangleright$ is not associative.

Part b: Testing $a \circledast b = a \times b$

Multiplication of real numbers is associative:

$(a \circledast b) \circledast c = (a \times b) \times c = abc$

$a \circledast (b \circledast c) = a \times (b \times c) = abc$

Since these are equal, $\circledast$ is associative.

Worked Example: Finding Identity and Self-Inverse Elements

Question: The binary operation $*$ is defined by $a * b = a + b$ on the set ℚ.

a) Find the identity element of the set ℚ under $*$.

b) Which element of ℚ is self-inverse under $*$?

Solution:

Part a: Finding the identity element

The identity element $e$ must satisfy $a * e = a$ for all $a$ in ℚ.

Using our operation: $a * e = a + e = a$

This means: $a + e = a$

Therefore: $e = 0$

We can verify: $a + 0 = 0 + a = a$ ✓

The identity element is 0.

Part b: Finding the self-inverse element

For an element $a$ to be self-inverse, we need $a * a = e$.

Since $e = 0$, we need: $a * a = 0$

Using our operation: $a + a = 0$

Therefore: $2a = 0$

So: $a = 0$

The only self-inverse element is 0.

Exam tip: When finding identity or inverse elements, always start with the defining equation and work algebraically. Remember that the identity element must work for all elements in the set.

Worked Example: Analyzing a Cayley Table

Question: For the Cayley table given:

a) Find the identity element.

b) Work out which of the elements are self-inverses.

Solution:

Part a: Finding the identity element

Looking at the fourth column (headed by $d$): $a * d = a$, $b * d = b$, $c * d = c$, $d * d = d$

This column is identical to the initial column of elements. We can verify by checking the fourth row, which also matches the initial row.

Therefore, $d$ is the identity element.

Part b: Finding self-inverse elements

We need to find all instances of $d$ (the identity) in the table, then check each row:

• In row $b$: $d$ appears in the column headed by $b$, so $b * b = d$, meaning $b$ is self-inverse
• In row $d$: $d$ appears in the column headed by $d$, so $d * d = d$, meaning $d$ is self-inverse

Also, examining the table:

• For element $a$: the inverse is $c$ (because $a * c = d$ and $c * a = d$)
• For element $c$: the inverse is $a$ (because $c * a = d$ and $a * c = d$)

Therefore, $b$ and $d$ are self-inverses, while $a$ and $c$ are inverses of each other.