DG11 - Generators of groups | AQA A-Level Further Mathematics

Generators build a group from a small collection of elements. You will list generated elements, identify generators of cyclic groups, and combine rotations and reflections to generate polygon symmetry groups. You will distinguish an individual generator from a generating set.

Generating elements by powers and products

Can two moves build all eight symmetries of a square, even though neither move can do so alone? A quarter-turn repeatedly produces rotations, and a reflection repeatedly produces only itself and the identity. Allowing products of both moves changes what is reachable.

Start with a single group element $a$ . Include $a$ , combine it with itself, include its inverse, and continue until no new elements appear. For a finite period $k$ , this produces exactly $e,a,a^2,\ldots,a^{k-1}$ ; the next power repeats $e$ . For infinite period, positive and negative powers never collapse to a finite list.

Generated subgroup and generator

The subgroup generated by $a$ is written

\langle a\rangle=\{a^m:m\in\mathbb Z\}

It contains all integer powers, including $a^0=e$ and negative powers of $a$ . If this is the whole underlying set of $G$ , then $a$ is a generator of $G$ .

The notation $G=(\langle n\rangle,*)$ says that $n$ generates $G$ under operation $*$ . Here $n$ denotes the generating element, not necessarily the group's order. The angle brackets denote generation, not the singleton set $\{n\}$ .

Under addition, powers mean repeated additions: $\langle3\rangle$ in $(\mathbb Z,+)$ is $\{\ldots,-6,-3,0,3,6,\ldots\}$ . Keep the operation visible, because the same written element can generate different sets under different operations.

Worked example 1: a generator under modular multiplication

Question. Let $G=(\{1,2,3,4,5,6\},\times_7)$ . Show that $3$ generates $G$ , and write $G$ using angle-bracket notation.

Plan. Multiply repeatedly by $3$ , reducing modulo $7$ , until the first return to $1$ .

\begin{array}{c|rrrrrr} m&1&2&3&4&5&6\\\hline 3^m\pmod7&3&2&6&4&5&1 \end{array}

For example, $3^3\equiv2\times3=6$ , then $3^4\equiv6\times3\equiv4\pmod7$ . Every element appears before repetition.

Answer. $3$ generates $G$ , so $G=(\langle3\rangle,\times_7)$ .

Check. Directly, $3^3=27\equiv-1\pmod7$ , so $3^6\equiv1$ . Also $3\times_7 5=1$ , agreeing with the fifth power being the inverse of $3$ . The six distinct listed outputs exhaust the six-element set.

Keep the group fixed and change only the candidate generator from $3$ to $2$ . A valid element need not generate the whole group.

Worked example 2: generating only a subgroup

Question. Find the subgroup generated by $2$ in the same group. Does $2$ generate $G$ ?

Plan. Repeat multiplication by $2$ until the first identity output, then compare with the full set.

2^1\equiv2,\qquad2^2\equiv4,\qquad2^3\equiv1\pmod7

Further multiplication repeats this cycle.

Answer. $(\langle2\rangle,\times_7)=(\{1,2,4\},\times_7)$ . The element $2$ does not generate $G$ , because $3,5,6$ are absent.

Check. Within this three-element subset, $2\times_7 2=4$ , $2\times_7 4=1$ , and $4\times_7 4=2$ . Thus products remain inside it; its inverse pair is $2\leftrightarrow4$ . No products can escape to produce a missing element.

A single element always generates a cyclic subgroup, even when that subgroup is proper.

Watch out

Reaching the identity shows that a cycle has closed. To call the starting element a generator of $G$ , check that the cycle contains every element of $G$ , not just some of them.

Try a fresh modulus and write both the set and the conclusion.

Generators of cyclic and polygon groups

In a cyclic group of order $n$ , choose a known generator $g$ . Repeating $g^k$ visits exponent positions $0,k,2k,\ldots$ modulo $n$ . A common factor of $k,n$ traps those positions in a smaller cycle.

Generators in a finite cyclic group

If $g$ generates a cyclic group of order $n$ , then

\text{period of }g^k=\frac{n}{\gcd(n,k)}

Consequently $g^k$ generates the group exactly when $\gcd(n,k)=1$ . Here $\gcd$ means greatest common divisor.

Not in the formula booklet: learn it.

To justify the fraction, write $n=dn'$ , $k=dk'$ , where $d=\gcd(n,k)$ and $n',k'$ have no common factor. The first return requires $n\mid kt$ , equivalently $n'\mid k't$ . Since no factor of $n'$ can come from $k'$ , the least positive $t$ is $n'$ . This explains the rule through the cycle rather than just supplying a test.

Worked example 3: a coprime step

Question. A cyclic group has order $10$ and generator $g$ . Determine whether $g^3$ is a generator.

Plan. Find the first return of the exponent step $3$ modulo $10$ .

\gcd(10,3)=1,\qquad\text{period of }g^3=\frac{10}{1}=10

Answer. Yes. $G=(\langle g^3\rangle,*)$ , where $*$ is the group operation.

Check. Successive exponent residues are $3,6,9,2,5,8,1,4,7,0$ . They visit all ten residues, independently confirming that every $g^j$ is reached.

Change only the exponent step, from $3$ to $4$ . A shared factor now prevents a full cycle.

Worked example 4: a step with a common factor

Question. In the same cyclic group, find $\langle g^4\rangle$ and determine whether $g^4$ generates $G$ .

Plan. Use the common factor to predict the cycle length, then list its exponent residues.

\gcd(10,4)=2,\qquad\text{period of }g^4=5

The powers have exponent residues $4,8,2,6,0$ .

Answer. $\langle g^4\rangle=\{e,g^2,g^4,g^6,g^8\}$ . It is a proper subgroup, so $g^4$ is not a generator of $G$ .

Check. Each exponent is even, and addition of even exponents stays even modulo $10$ . In particular, $g$ cannot appear; the five distinct even residues show that the list is complete.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Core Pure

Mechanics

Statistics

Discrete Mathematics

DG11 - Generators of groups

Generating elements by powers and products

Generators of cyclic and polygon groups

Continue this lesson.

Learn faster with Nova