## Question

Let *G* be a finite cyclic group.

(a) Prove that *G* is Abelian.

(b) Given that *a* is a generator of *G*, show that \({a^{ – 1}}\) is also a generator.

(c) Show that if the order of *G* is five, then all elements of *G*, apart from the identity, are generators of *G*.

**Answer/Explanation**

## Markscheme

(a) let *a* be a generator and consider the (general) elements \(b = {a^m},{\text{ }}c = {a^n}\) *M1*

then

\(bc = {a^m}{a^n}\) *A1*

\( = {a^n}{a^m}\) (using associativity) *R1*

\( = cb\) *A1*

therefore *G* is Abelian *AG*

*[4 marks]*

* *

(b) let *G* be of order *p* and let \(m \in \{ 1,…….,{\text{ }}p\} \), let *a* be a generator

consider \(a{a^{ – 1}} = e \Rightarrow {a^m}{({a^{ – 1}})^m} = e\) *M1R1*

this shows that \({({a^{ – 1}})^m}\) is the inverse of \({a^m}\) *R1*

as *m* increases from 1 to *p*, \({a^m}\) takes *p* different values and it generates *G* *R1*

it follows from the uniqueness of the inverse that \({({a^{ – 1}})^m}\) takes *p* different values and is a generator *R1*

*[5 marks]*

* *

(c) **EITHER**

by Lagrange, the order of any element divides the order of the group, *i.e.* 5 *R1*

the only numbers dividing 5 are 1 and 5 *R1*

the identity element is the only element of order 1 *R1*

all the other elements must be of order 5 *R1*

so they all generate *G* *AG*

**OR**

let *a* be a generator.

successive powers of *a* and therefore the elements of *G* are

\(a,{\text{ }}{a^2},{\text{ }}{a^3},{\text{ }}{a^4}{\text{ and }}{a^5} = e\) *A1*

successive powers of \({a^2}\) are \({a^2},{\text{ }}{a^4},{\text{ }}a,{\text{ }}{a^3},{\text{ }}{a^5} = e\) *A1*

successive powers of \({a^3}\) are \({a^3},{\text{ }}a,{\text{ }}{a^4},{\text{ }}{a^2},{\text{ }}{a^5} = e\) *A1*

successive powers of \({a^4}\) are \({a^4},{\text{ }}{a^3},{\text{ }}{a^2},{\text{ }}a,{\text{ }}{a^5} = e\) *A1*

this shows that \({a^2},{\text{ }}{a^3},{\text{ }}{a^4}\) are also generators in addition to *a* *AG*

*[4 marks]*

*Total [13 marks]*

## Examiners report

Solutions to (a) were often disappointing with some solutions even stating that a cyclic group is, by definition, commutative and therefore Abelian. Explanations in (b) were often poor and it was difficult in some cases to distinguish between correct and incorrect solutions. In (c), candidates who realised that Lagrange’s Theorem could be used were generally the most successful. Solutions again confirmed that, in general, candidates find theoretical questions on this topic difficult.