Subgroups (AQA A-Level Further Maths): Revision Notes

Worked Example 1a: Proving a Subgroup

Show that $H = (\mathbb{Q}, +)$ is a subgroup of $G = (\mathbb{R}, +)$

Note: $\mathbb{R} \backslash \{0\}$ denotes the set of real numbers excluding zero.

Solution:

We check each condition systematically:

Non-empty: $H$ is non-empty since there exist infinitely many rational numbers (for instance, $1 \in \mathbb{Q}$).

Identity: The identity element of $G$ under addition is $0$, and $0 \in \mathbb{Q}$.

Closure: If $x, y \in \mathbb{Q}$, then $x + y \in \mathbb{Q}$ (the sum of two rational numbers is rational).

Inverses: The inverse of an element $x$ under addition is $-x$. If $x \in \mathbb{Q}$, then $-x \in \mathbb{Q}$.

Since all four conditions are satisfied, $H$ is a subgroup of $G$.

Worked Example 1b: Disproving a Subgroup Using Counter-Example

Show that the set $J = \{x \in \mathbb{R} : x \geq 1\}$ is not a subgroup of $K = (\mathbb{R} \backslash \{0\}, \times)$

Solution:

We need only find one condition that fails:

Non-empty: $J$ is non-empty since there exist infinitely many real numbers greater than or equal to 1. ✓

Identity: The identity element of $K$ under multiplication is $1$, and $1 \in J$. ✓

Closure: If $x$ and $y$ are in $J$, then their product $xy$ is in $J$ (multiplying numbers $\geq 1$ gives a result $\geq 1$). ✓

Inverses: The inverse of an element $x$ under multiplication is $x^{-1} = \frac{1}{x}$.

However, if $x > 1$, then $x^{-1} = \frac{1}{x} < 1$, so $x^{-1} \notin J$. ✗

Since $J$ does not contain the inverses of all its elements, $J$ is not a subgroup of $K$.

Worked Example 2: Finding Subgroups Using a Cayley Table

A group $G$ is formed by the set $S = \{a, b, c, d\}$ under the binary operation $\bullet$.

The Cayley table shows the result of each operation:

Task: Find all the non-trivial subgroups of $G$.

Solution:

Step 1: First, identify the identity element. From the Cayley table, $a$ is the identity element since $a \bullet x = x \bullet a = x$ for all elements.

Step 2: Since we want non-trivial subgroups, we exclude $\{a\}$ (the trivial subgroup). Since subgroups must operate under the same binary operation as the original group, all subgroups must contain the identity element $a$.

Step 3: The possible subsets containing $a$ are: $\{a, b\}$, $\{a, c\}$, $\{a, d\}$, $\{a, b, c\}$, $\{a, b, d\}$, $\{a, c, d\}$ and $\{a, b, c, d\}$ (the group itself).

Now we check each subset systematically:

• $\{a, b\}$: We check closure. Since $b \bullet b = a$ (from the table), and both $a$ and $b$ are self-inverse, $\{a, b\}$ is closed. Therefore, $\{a, b\}$ is a subgroup.

• $\{a, c\}$: Check closure. From the table, $c \bullet c = b$, but $b \notin \{a, c\}$. Not closed, so not a subgroup.

• $\{a, d\}$: Check closure. From the table, $d \bullet d = b$, but $b \notin \{a, d\}$. Not closed, so not a subgroup.

• $\{a, b, c\}$: Check closure. From the table, $b \bullet c = d$, but $d \notin \{a, b, c\}$. Not closed, so not a subgroup.

• $\{a, b, d\}$: Check closure. From the table, $b \bullet d = c$, but $c \notin \{a, b, d\}$. Not closed, so not a subgroup.

• $\{a, c, d\}$: Check closure. From the table, $c \bullet d = b$, but $b \notin \{a, c, d\}$. Not closed, so not a subgroup.

• $\{a, b, c, d\}$: This is the group itself, so by definition it is a non-trivial subgroup.

Conclusion: The non-trivial subgroups of $G$ are $\{a, b\}$ and $\{a, b, c, d\}$.

Worked Example 3: Using Lagrange's Theorem to Find Subgroups

The group $G = (\{0, 1, 2, 3, 4, 5\}, +_6)$ uses addition modulo 6.

Part a: State the order of $G$.

Solution: There are six elements in $G$, so the order is 6.

Part b: Write down the order of possible subgroups of $G$. Justify your answer.

Solution: By Lagrange's theorem, the order of any subgroup must divide the order of $G$.

The factors of 6 are 1, 2, 3, and 6.

Therefore, possible subgroup orders are 1, 2, 3, and 6.

Part c: Use your answer to part b to find all the proper subgroups of $G$.

Solution:

We list all possibilities based on the orders found in part b:

• Order 1: $\{0\}$ is the trivial subgroup (not proper if we consider the definition strictly, but typically we seek non-trivial proper subgroups).

• Order 6: $\{0, 1, 2, 3, 4, 5\}$ is the group itself, so it is not a proper subgroup.

• Order 2: For two-element subgroups, the element other than the identity must be self-inverse.

  Checking: $2 + 2 \equiv 4 \pmod{6}$, $3 + 3 \equiv 0 \pmod{6}$, $4 + 4 \equiv 2 \pmod{6}$, $5 + 5 \equiv 4 \pmod{6}$.

  Only 3 is self-inverse (besides 0), so $\{0, 3\}$ is a subgroup.

• Order 3: For three-element subgroups, both elements other than the identity must be inverses of each other, or they must be self-inverse.

  Since 3 is the only self-inverse element other than the identity, we consider sets like $\{0, 1, 2\}$, $\{0, 2, 3\}$, $\{0, 3, 4\}$, and $\{0, 3, 5\}$.

  Testing closure for inverses:

  • $2 + 4 \equiv 0 \pmod{6}$ and $1 + 5 \equiv 0 \pmod{6}$

  For $\{0, 1, 5\}$: Need to check if closed. $1 + 1 = 2 \notin \{0, 1, 5\}$. Not closed.

  For $\{0, 2, 4\}$: Check closure. $2 + 2 = 4$ ✓, $2 + 4 = 0$ ✓, $4 + 4 = 2$ ✓. This is closed, so $\{0, 2, 4\}$ is a subgroup.

Answer: The proper subgroups of $G$ are $\{0, 3\}$ and $\{0, 2, 4\}$ (and the trivial subgroup $\{0\}$ if we include it).