DG7, DG8 - The group axioms | AQA A-Level Further Mathematics

A group is a set with an operation that combines and undoes elements consistently. You will prove the four group axioms, distinguish a group's order from an element's period, and use Cayley tables while recognising what a table pattern does not prove.

Proving the four group axioms

Why do the residues $\{0,1,2,3\}$ form a group under addition modulo $4$ , but not under multiplication modulo $4$ ? Both rules are defined and closed. The difference appears when you try to undo an element.

Group axioms

A group $(S,*)$ is a nonempty set $S$ with an operation satisfying all four conditions:

Closure: $a*b\in S$ for every $a,b\in S$ , with the output uniquely defined.
Identity: there is $e\in S$ with $e*a=a*e=a$ for every $a\in S$ .
Inverses: for every $a\in S$ , there is $b\in S$ with $a*b=b*a=e$ .
Associativity: $(a*b)*c=a*(b*c)$ for every $a,b,c\in S$ .

Commutativity is not a group axiom. Noncommutative multiplication can still form a group; proving that inputs cannot be swapped does not disprove a group claim.

To prove a group, establish every axiom for general allowed inputs. To disprove a group, one failed axiom is enough, but explain exactly why it fails. If a question asks you to consider a particular axiom, address that demand.

Worked example 1: a complete group proof

Question. Prove that $\mathbb Z$ forms a group under $a*b=a+b-4$ .

Plan. Check closure, find an allowed identity and general inverse, then compare both bracketings.

Closure. If $a,b\in\mathbb Z$ , then $a+b-4\in\mathbb Z$ .

Identity. $e=4\in\mathbb Z$ , since

a*4=a+4-4=a,\qquad4*a=4+a-4=a.

Inverses. For any $a\in\mathbb Z$ , choose $b=8-a\in\mathbb Z$ . Then

a*(8-a)=a+8-a-4=4,\qquad(8-a)*a=4.

Associativity. For arbitrary $a,b,c\in\mathbb Z$ ,

(a*b)*c=a+b+c-8=a*(b*c).

Answer. All four axioms hold, so $(\mathbb Z,*)$ is a group.

Check. Subtract $4$ from each element. The output becomes $(a*b)-4=(a-4)+(b-4)$ , agreeing with the identity $4$ and inverse $8-a$ found above.

Keep the rule fixed, but restrict the set. Closure and an identity can survive while the inverse axiom fails.

Worked example 2: one failed axiom

Question. Does $S=\{4,5,6,\ldots\}$ form a group under $a*b=a+b-4$ ?

Plan. Look for an element whose required inverse is excluded.

Closure holds because $a,b\ge4$ gives $a+b-4\ge4$ . The identity $4$ is allowed, and the same associativity proof still applies. But an inverse of $5$ must satisfy

5*b=4\quad\Longrightarrow\quad5+b-4=4\quad\Longrightarrow\quad b=3.

Answer. This is not a group: $3\notin S$ , so $5$ has no inverse in $S$ .

Check. For every allowed $b\ge4$ , $5*b=b+1\ge5$ , which cannot be the identity $4$ . This directly excludes every possible inverse candidate.

The noncommutative pair operation from the preceding lesson also gives a complete group example. On $S=\{(a,b):a\in\mathbb R,\ b\in\mathbb R\setminus\{0\}\}$ , set

(a,b)\circ(c,d)=(a+bc,bd).

Closure holds because $bd\ne0$ ; the identity is $(0,1)$ , and the two-sided inverse of $(a,b)$ is $(-a/b,1/b)$ , which is allowed. For a general third pair $(f,g)$ , both bracketings give

\begin{aligned} ((a,b)\circ(c,d))\circ(f,g)&=(a+bc+bdf,bdg)\\ (a,b)\circ((c,d)\circ(f,g))&=(a+b(c+df),b(dg))\\ &=(a+bc+bdf,bdg). \end{aligned}

Together with the two-sided checks already proved, this establishes all axioms. Yet $(1,2)\circ(0,3)=(1,6)$ and $(0,3)\circ(1,2)=(3,6)$ : the group is noncommutative.

Watch out

A group proof needs all four axioms, with the quantifiers “every pair”, “every element” and “every triple” respected. A no-group conclusion needs a specific failed axiom, not a missing commutative law.

Use a new shifted operation and provide every part of the proof.

Group order and element period

The size of a group and the number of repetitions needed to return to its identity measure different things. A group can have six elements while a particular element returns to the identity after only two repetitions.

Order and period

The order of a finite group is its number of elements, written $|G|$ .

The period of an element $a$ is the least positive integer $k$ for which combining $a$ with itself $k$ times gives the identity. This is often written $a^k=e$ , where powers mean repetition of the group's operation. If no such positive integer exists, the element has infinite period.

The identity has period $1$ , not $0$ : “least positive” is part of the definition. Under an additive operation, repeated combination is usually written $ka$ , not ordinary exponentiation. Under $+_n$ , return to the identity means remainder $0$ ; under $\times_n$ , it means remainder $1$ .

Worked example 3: finding a period by repetition

Question. In $G=(\{0,1,2,3,4,5\},+_6)$ , state the group order and find the period of $2$ .

Plan. Count elements, then list repeated additions until the first return to $0$ .

\begin{array}{c|ccc} \text{number of copies of }2&1&2&3\\\hline \text{output modulo }6&2&4&0 \end{array}

The outputs at one and two copies are not the identity. Three copies give $2+2+2=6\equiv0\pmod6$ .

Answer. $|G|=6$ , and $2$ has period $3$ .

Check. A positive number of copies $k$ works exactly when $2k$ is divisible by $6$ . The smallest such $k$ is $3$ , confirming minimality.

Keep the group fixed and change only the element from $2$ to $3$ . Its period need not stay the same.

Worked example 4: a different element in the same group

Question. Find the period of $3$ in the same group.

Plan. Start the repetition count again; do not use the previous element's period.

\begin{array}{c|cc} \text{number of copies of }3&1&2\\\hline \text{output modulo }6&3&0 \end{array}

Answer. The period of $3$ is $2$ . The group still has order $6$ .

Check. $3\ne0$ , so period $1$ is impossible; $3+3=6\equiv0\pmod6$ , so $2$ is the first possible positive return.

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DG7, DG8 - The group axioms

Proving the four group axioms

Group order and element period

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