# IB DP Maths Topic 8.11 Subgroups, proper subgroups HL Paper 3

## Question

(a)     Draw the Cayley table for the set of integers G = {0, 1, 2, 3, 4, 5} under addition modulo 6, $${ + _6}$$.

(b)     Show that $$\{ G,{\text{ }}{ + _6}\}$$ is a group.

(c)     Find the order of each element.

(d)     Show that $$\{ G,{\text{ }}{ + _6}\}$$ is cyclic and state its generators.

(e)     Find a subgroup with three elements.

(f)     Find the other proper subgroups of $$\{ G,{\text{ }}{ + _6}\}$$.

## Markscheme

(a)          A3

Note: Award A2 for 1 error, A1 for 2 errors and A0 for more than 2 errors.

[3 marks]

(b)     The table is closed     A1

Identity element is 0     A1

Each element has a unique inverse (0 appears exactly once in each row and column)     A1

Addition mod 6 is associative     A1

Hence $$\{ G,{\text{ }}{ + _6}\}$$ forms a group     AG

[4 marks]

(c)     0 has order 1 (0 = 0),

1 has order 6 (1 + 1 + 1 + 1 + 1 + 1 = 0),

2 has order 3 (2 + 2 + 2 = 0),

3 has order 2 (3 + 3 = 0),

4 has order 3 (4 + 4 + 4 = 0),

5 has order 6 (5 + 5 + 5 + 5 + 5 + 5 = 0).     A3

Note: Award A2 for 1 error, A1 for 2 errors and A0 for more than 2 errors.

[3 marks]

(d)     Since 1 and 5 are of order 6 (the same as the order of the group) every element can be written as sums of either 1 or 5. Hence the group is cyclic.     R1

The generators are 1 and 5.     A1

[2 marks]

(e)     A subgroup of order 3 is $$\left( {\{ 0,{\text{ }}2,{\text{ }}4\} ,{\text{ }}{ + _6}} \right)$$     A2

Note: Award A1 if only {0, 2, 4} is seen.

[2 marks]

(f)     Other proper subgroups are $$\left( {\{ 0\} { + _6}} \right),{\text{ }}\left( {\{ 0,{\text{ }}3\} { + _6}} \right)$$     A1A1

Note: Award A1 if only {0}, {0, 3} is seen.

[2 marks]

Total [16 marks]

## Examiners report

The table was well done as was showing its group properties. The order of the elements in (b) was done well except for the order of 0 which was often not given. Finding the generators did not seem difficult but correctly stating the subgroups was not often done. The notion of a ‘proper’ subgroup is not well known.

## Question

The relation aRb is defined on {1, 2, 3, 4, 5, 6, 7, 8, 9} if and only if ab is the square of a positive integer.

(i)     Show that R is an equivalence relation.

(ii)     Find the equivalence classes of R that contain more than one element.

[10]
a.

Given the group $$(G,{\text{ }} * )$$, a subgroup $$(H,{\text{ }} * )$$ and $$a,{\text{ }}b \in G$$, we define $$a \sim b$$ if and only if $$a{b^{ – 1}} \in H$$. Show that $$\sim$$ is an equivalence relation.

[9]
b.

## Markscheme

(i)     $$aRa \Rightarrow a \cdot a = {a^2}$$ so R is reflexive     A1

$$aRb = {m^2} \Rightarrow bRa$$ so R is symmetric     A1

$$aRb = ab = {m^2}{\text{ and }}bRc = bc = {n^2}$$     M1A1

so $$a = \frac{{{m^2}}}{b}{\text{ and }}c = \frac{{{n^2}}}{b}$$

$$ac = \frac{{{m^2}{n^2}}}{{{b^2}}} = {\left( {\frac{{mn}}{b}} \right)^2},$$     A1

ac is an integer hence $${\left( {\frac{{mn}}{b}} \right)^2}$$ is an integer     R1

so aRc, hence R is transitive     R1

R is therefore an equivalence relation     AG

(ii)     1R4 and 4R9 or 2R8     M1

so {1, 4, 9} is an equivalence class     A1

and {2, 8} is an equivalence class     A1

[10 marks]

a.

$$a \sim a{\text{ since }}a{a^{ – 1}} = e \in H$$, the identity must be in H since it is a subgroup.     M1

Hence reflexivity.     R1

$$a \sim b \Leftrightarrow a{b^{ – 1}} \in H$$ but H is a subgroup so it must contain $${(a{b^{ – 1}})^{ – 1}} = b{a^{ – 1}}$$     M1R1

i.e. $$b{a^{ – 1}} \in H{\text{ so }} \sim$$ is symmetric     A1

$$a \sim b{\text{ and }}b \sim c \Rightarrow a{b^{ – 1}} \in H{\text{ and }}b{c^{ – 1}} \in H$$     M1

But H is closed, so

$$(a{b^{ – 1}})(b{c^{ – 1}}) \in H{\text{ or }}a({b^{ – 1}}b){c^{ – 1}} \in H$$     R1

$$a{c^{ – 1}} \in H \Rightarrow a \sim c$$     A1

Hence $$\sim$$ is transitive and is thus an equivalence relation     R1AG

[9 marks]

b.

## Examiners report

Not a difficult question although using the relation definition to fully show transitivity was not well done. It was good to see some students use an operation binary matrix to show transitivity. This was a nice way given that the set was finite. The proof in (b) proved difficult.

a.

Not a difficult question although using the relation definition to fully show transitivity was not well done. It was good to see some students use an operation binary matrix to show transitivity. This was a nice way given that the set was finite. The proof in (b) proved difficult.

b.

## Question

Let {G , $$*$$} be a finite group of order n and let H be a non-empty subset of G .

(a)     Show that any element $$h \in H$$ has order smaller than or equal to n .

(b)     If H is closed under $$*$$, show that {H , $$*$$} is a subgroup of {G , $$*$$}.

## Markscheme

(a)     if $$h \in H$$ then $$h \in G$$     R1

hence, (by Lagrange) the order of h exactly divides n

and so the order of h is smaller than or equal to n     R2

[3 marks]

(b)     the associativity in G ensures associativity in H     R1

(closure within H is given)

as H is non-empty there exists an $$h \in H$$ , let the order of h be m then $${h^m} = e$$ and as H is closed $$e \in H$$     R2

it follows from the earlier result that $$h * {h^{m – 1}} = {h^{m – 1}} * h = e$$     R1

thus, the inverse of h is $${h^{m – 1}}$$ which $$\in H$$     R1

the four axioms are satisfied showing that $$\{ H{\text{ , }} * \}$$ is a subgroup     R1

[6 marks]

Total [9 marks]

## Examiners report

Solutions to this question were extremely disappointing. This property of subgroups is mentioned specifically in the Guide and yet most candidates were unable to make much progress in (b) and even solutions to (a) were often unconvincing.

## Question

Set $$S = \{ {x_0},{\text{ }}{x_1},{\text{ }}{x_2},{\text{ }}{x_3},{\text{ }}{x_4},{\text{ }}{x_5}\}$$ and a binary operation $$\circ$$ on S is defined as $${x_i} \circ {x_j} = {x_k}$$, where $$i + j \equiv k(\bmod 6)$$.

(a)     (i)     Construct the Cayley table for $$\{ S,{\text{ }} \circ \}$$ and hence show that it is a group.

(ii)     Show that $$\{ S,{\text{ }} \circ \}$$ is cyclic.

(b)     Let $$\{ G,{\text{ }} * \}$$ be an Abelian group of order 6. The element $$a \in {\text{G}}$$ has order 2 and the element $$b \in {\text{G}}$$ has order 3.

(i)     Write down the six elements of $$\{ G,{\text{ }} * \}$$.

(ii)     Find the order of $${\text{a}} * b$$ and hence show that $$\{ G,{\text{ }} * \}$$ is isomorphic to $$\{ S,{\text{ }} \circ \}$$.

## Markscheme

(a)     (i)     Cayley table for $$\{ S,{\text{ }} \circ \}$$

$$\begin{array}{*{20}{c|cccccc}} \circ &{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}} \\ \hline {{x_0}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}} \\ {{x_1}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}} \\ {{x_2}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}} \\ {{x_3}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}} \\ {{x_4}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}} \\ {{x_5}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}} \end{array}$$     A4

Note: Award A4 for no errors, A3 for one error, A2 for two errors, A1 for three errors and A0 for four or more errors.

S is closed under $$\circ$$     A1

$${x_0}$$ is the identity     A1

$${x_0}$$ and $${x_3}$$ are self-inverses,     A1

$${x_2}$$ and $${x_4}$$ are mutual inverses and so are $${x_1}$$ and $${x_5}$$     A1

hence, $$\{ S,{\text{ }} \circ \}$$ is a group     AG

(ii)     the order of $${x_1}$$ (or $${x_5}$$) is 6, hence there exists a generator, and $$\{ S,{\text{ }} \circ \}$$ is a cyclic group     A1R1

[11 marks]

(b)     (i)     e, a, b, ab     A1

and $${b^2},{\text{ }}a{b^2}$$     A1A1

Note: Accept $$ba$$ and $${b^2}a$$.

(ii)     $${(ab)^2} = {b^2}$$     M1A1

$${(ab)^3} = a$$     A1

$${(ab)^4} = b$$     A1

hence order is 6     A1

groups G and S have the same orders and both are cyclic     R1

hence isomorphic     AG

[9 marks]

Total [20 marks]

## Examiners report

a) Most candidates had the correct Cayley table and were able to show successfully that the group axioms were satisfied. Some candidates, however, simply stated that an inverse exists for each element without stating the elements and their inverses. Most candidates were able to find a generator and hence show that the group is cyclic.

b) This part was answered less successfully by many candidates. Some failed to find all the elements. Some stated that the order of ab is 6 without showing any working.

## Question

The binary operator multiplication modulo 14, denoted by $$*$$, is defined on the set S = {2, 4, 6, 8, 10, 12}.

Copy and complete the following operation table.

[4]
a.

(i)     Show that {S , $$*$$} is a group.

(ii)     Find the order of each element of {S , $$*$$}.

(iii)     Hence show that {S , $$*$$} is cyclic and find all the generators.

[11]
b.

The set T is defined by $$\{ x * x:x \in S\}$$. Show that {T , $$*$$} is a subgroup of {S , $$*$$}.

[3]
c.

## Markscheme

A4

Note: Award A4 for all correct, A3 for one error, A2 for two errors, A1 for three errors and A0 for four or more errors.

[4 marks]

a.

(i)     closure: there are no new elements in the table     A1

identity: 8 is the identity element     A1

inverse: every element has an inverse because there is an 8 in every row and column     A1

associativity: (modulo) multiplication is associative     A1

therefore {S , $$*$$} is a group     AG

(ii)     the orders of the elements are as follows

A4

Note: Award A4 for all correct, A3 for one error, A2 for two errors, A1 for three errors and A0 for four or more errors.

(iii)     EITHER

the group is cyclic because there are elements of order 6     R1

OR

the group is cyclic because there are generators     R1

THEN

10 and 12 are the generators     A1A1

[11 marks]

b.

looking at the Cayley table, we see that

T = {2, 4, 8}     A1

this is a subgroup because it contains the identity element 8, no new elements are formed and 2 and 4 form an inverse pair     R2

Note: Award R1 for any two conditions

[3 marks]

c.

## Examiners report

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of T. This approach was invariably unsuccessful.

a.

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of T. This approach was invariably unsuccessful.

b.

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of T. This approach was invariably unsuccessful.

c.

## Question

Consider the following Cayley table for the set G = {1, 3, 5, 7, 9, 11, 13, 15} under the operation $${ \times _{16}}$$, where $${ \times _{16}}$$ denotes multiplication modulo 16.

(i)     Find the values of a, b, c, d, e, f, g, h, i and j.

(ii)     Given that $${ \times _{16}}$$ is associative, show that the set G, together with the operation $${ \times _{16}}$$, forms a group.

[7]
a.

The Cayley table for the set $$H = \{ e,{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }}{b_1},{\text{ }}{b_2},{\text{ }}{b_3},{\text{ }}{b_4}\}$$ under the operation $$*$$, is shown below.

(i)     Given that $$*$$ is associative, show that H together with the operation $$*$$ forms a group.

(ii)     Find two subgroups of order 4.

[8]
b.

Show that $$\{ G,{\text{ }}{ \times _{16}}\}$$ and $$\{ H,{\text{ }} * \}$$ are not isomorphic.

[2]
c.

Show that $$\{ H,{\text{ }} * \}$$ is not cyclic.

[3]
d.

## Markscheme

(i)     $$a = 9,{\text{ }}b = 1,{\text{ }}c = 13,{\text{ }}d = 5,{\text{ }}e = 15,{\text{ }}f = 11,{\text{ }}g = 15,{\text{ }}h = 1,{\text{ }}i = 15,{\text{ }}j = 15$$     A3

Note: Award A2 for one or two errors,

A1 for three or four errors,

A0 for five or more errors.

(ii)     since the Cayley table only contains elements of the set G, then it is closed     A1

there is an identity element which is 1     A1

{3, 11} and {5, 13} are inverse pairs and all other elements are self inverse     A1

hence every element has an inverse     R1

Note: Award A0R0 if no justification given for every element having an inverse.

since the set is closed, has an identity element, every element has an inverse and it is associative, it is a group     AG

[7 marks]

a.

(i)     since the Cayley table only contains elements of the set H, then it is closed     A1

there is an identity element which is e     A1

$$\{ {a_1},{\text{ }}{a_3}\}$$ form an inverse pair and all other elements are self inverse     A1

hence every element has an inverse     R1

Note: Award A0R0 if no justification given for every element having an inverse.

since the set is closed, has an identity element, every element has an inverse and it is associative, it is a group     AG

(ii)     any 2 of $$\{ e,{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }}{a_3}\} ,{\text{ }}\{ e,{\text{ }}{a_2},{\text{ }}{b_1},{\text{ }}{b_2}\} ,{\text{ }}\{ e,{\text{ }}{a_2},{\text{ }}{b_3},{\text{ }}{b_4}\}$$     A2A2

[8 marks]

b.

the groups are not isomorphic because $$\{ H,{\text{ }} * \}$$ has one inverse pair whereas $$\{ G,{\text{ }}{ \times _{16}}\}$$ has two inverse pairs     A2

Note: Accept any other valid reason:

e.g. the fact that $$\{ G,{\text{ }}{ \times _{16}}\}$$ is commutative and $$\{ H,{\text{ }} * \}$$ is not.

[2 marks]

c.

EITHER

a group is not cyclic if it has no generators     R1

for the group to have a generator there must be an element in the group of order eight     A1

since there is no element of order eight in the group, it is not cyclic     A1

OR

a group is not cyclic if it has no generators     R1

only possibilities are $${a_1}$$, $${a_3}$$ since all other elements are self inverse     A1

this is not possible since it is not possible to generate any of the “b” elements from the “a” elements – the elements $${a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }}{a_4}$$ form a closed set     A1

[3 marks]

d.

## Examiners report

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

a.

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

b.

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

c.

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

d.

## Question

Associativity and commutativity are two of the five conditions for a set S with the binary operation $$*$$ to be an Abelian group; state the other three conditions.

[2]
a.

The Cayley table for the binary operation $$\odot$$ defined on the set T = {p, q, r, s, t} is given below.

(i)     Show that exactly three of the conditions for {T , $$\odot$$} to be an Abelian group are satisfied, but that neither associativity nor commutativity are satisfied.

(ii)     Find the proper subsets of T that are groups of order 2, and comment on your result in the context of Lagrange’s theorem.

(iii)     Find the solutions of the equation $$(p \odot x) \odot x = x \odot p$$ .

[15]
b.

## Markscheme

closure, identity, inverse     A2

Note: Award A1 for two correct properties, A0 otherwise.

[2 marks]

a.

(i)     closure: there are no extra elements in the table     R1

identity: s is a (left and right) identity     R1

inverses: all elements are self-inverse     R1

commutative: no, because the table is not symmetrical about the leading diagonal, or by counterexample     R1

associativity: for example, $$(pq)t = rt = p$$     M1A1

not associative because $$p(qt) = pr = t \ne p$$     R1

Note: Award M1A1 for 1 complete example whether or not it shows non-associativity.

(ii)     $$\{ s,\,p\} ,{\text{ }}\{ s,\,q\} ,{\text{ }}\{ s,\,r\} ,{\text{ }}\{ s,\,t\}$$     A2

Note: Award A1 for 2 or 3 correct sets.

as 2 does not divide 5, Lagrange’s theorem would have been contradicted if T had been a group     R1

(iii)     any attempt at trying values     (M1)

the solutions are q, r, s and t     A1A1A1A1

Note: Deduct A1 if p is included.

[15 marks]

b.

## Examiners report

This was on the whole a well answered question and it was rare for a candidate not to obtain full marks on part (a). In part (b) the vast majority of candidates were able to show that the set satisfied the properties of a group apart from associativity which they were also familiar with. Virtually all candidates knew the difference between commutativity and associativity and were able to distinguish between the two. Candidates were familiar with Lagrange’s Theorem and many were able to see how it did not apply in the case of this problem. Many candidates found a solution method to part (iii) of the problem and obtained full marks.

a.

This was on the whole a well answered question and it was rare for a candidate not to obtain full marks on part (a). In part (b) the vast majority of candidates were able to show that the set satisfied the properties of a group apart from associativity which they were also familiar with. Virtually all candidates knew the difference between commutativity and associativity and were able to distinguish between the two. Candidates were familiar with Lagrange’s Theorem and many were able to see how it did not apply in the case of this problem. Many candidates found a solution method to part (iii) of the problem and obtained full marks.

b.

## Question

Consider the set S = {1, 3, 5, 7, 9, 11, 13} under the binary operation multiplication modulo 14 denoted by $${ \times _{14}}$$.

Copy and complete the following Cayley table for this binary operation.

[4]
a.

Give one reason why $$\{ S,{\text{ }}{ \times _{14}}\}$$ is not a group.

[1]
b.

Show that a new set G can be formed by removing one of the elements of S such that $$\{ G,{\text{ }}{ \times _{14}}\}$$ is a group.

[5]
c.

Determine the order of each element of $$\{ G,{\text{ }}{ \times _{14}}\}$$.

[4]
d.

Find the proper subgroups of $$\{ G,{\text{ }}{ \times _{14}}\}$$.

[2]
e.

## Markscheme

A4

Note: Award A3 for one error, A2 for two errors, A1 for three errors, A0 for four or more errors.

[4 marks]

a.

any valid reason, for example     R1

not a Latin square

7 has no inverse

[1 mark]

b.

delete 7 (so that G = {1, 3, 5, 9, 11, 13})     A1

closure – evident from the table     A1

associative because multiplication is associative     A1

the identity is 1     A1

13 is self-inverse, 3 and 5 form an inverse

pair and 9 and 11 form an inverse pair     A1

the four conditions are satisfied so that $$\{ G,{\text{ }}{ \times _{14}}\}$$ is a group     AG

[5 marks]

c.

A4

Note: Award A3 for one error, A2 for two errors, A1 for three errors, A0 for four or more errors.

[4 marks]

d.

{1}

{1, 13}$$\,\,\,\,\,$${1, 9, 11}     A1A1

[2 marks]

e.

## Examiners report

There were no problems with parts (a), (b) and (d).

a.

There were no problems with parts (a), (b) and (d).

b.

There were no problems with parts (a), (b) and (d) but in part (c) candidates often failed to state that the set was associative under the operation because multiplication is associative. Likewise they often failed to list the inverses of each element simply stating that the identity was present in each row and column of the Cayley table.

c.

The majority of candidates did not answer part (d) correctly and often simply listed all subsets of order 2 and 3 as subgroups.

d.

[N/A]

e.

## Question

H and K are subgroups of a group G. By considering the four group axioms, prove that $$H \cap K$$ is also a subgroup of G.

## Markscheme

closure: let $$a,{\text{ }}b \in H \cap K$$, so that $$a,{\text{ }}b \in H$$ and $$a,{\text{ }}b \in K$$     M1

therefore $$ab \in H$$ and $$ab \in K$$ so that $$ab \in H \cap K$$     A1

associativity: this carries over from G     R1

identity: the identity $$e \in H$$ and $$e \in K$$     M1

therefore $$e \in H \cap K$$     A1

inverse:

$$a \in H \cap K$$ implies $$a \in H$$ and $$a \in K$$     M1

it follows that $${a^{ – 1}} \in H$$ and $${a^{ – 1}} \in K$$     A1

and therefore that $${a^{ – 1}} \in H \cap K$$     A1

the four group axioms are therefore satisfied     AG

[8 marks]

## Examiners report

This question presented the most difficulty for students. Overall the candidates showed a lack of ability to present a formal proof. Some gained points for the proof of the identity element in the intersection and the statement that the associative property carries over from the group. However, the vast majority gained no points for the proof of closure or the inverse axioms.

## Question

Let $$G$$ be a group of order 12 with identity element e.

Let $$a \in G$$ such that $${a^6} \ne e$$ and $${a^4} \ne e$$.

(i)      Prove that $$G$$ is cyclic and state two of its generators.

(ii)     Let $$H$$ be the subgroup generated by $${a^4}$$. Construct a Cayley table for $$H$$.

[9]
a.

State, with a reason, whether or not it is necessary that a group is cyclic given that all its proper subgroups are cyclic.

[2]
b.

## Markscheme

(i)   the order of $$a$$ is a divisor of the order of $$G$$     (M1)

since the order of $$G$$ is 12, the order of $$a$$ must be 1, 2, 3, 4, 6 or 12     A1

the order cannot be 1, 2, 3 or 6, since $${a^6} \ne e$$     R1

the order cannot be 4, since $${a^4} \ne e$$     R1

so the order of $$a$$ must be 12

therefore, $$a$$ is a generator of $$G$$, which must therefore be cyclic     R1

another generator is eg $${a^{ – 1}},{\text{ }}{a^5},{\text{ }} \ldots$$     A1

[6 marks]

(ii)     $$H = \{ e,{\text{ }}{a^4},{\text{ }}{a^8}\}$$     (A1)

M1A1

[3 marks]

a.

no     A1

eg the group of symmetries of a triangle $${S_3}$$ is not cyclic but all its (proper) subgroups are cyclic

eg the Klein four-group is not cyclic but all its (proper) subgroups are cyclic     R1

[2 marks]

b.

## Examiners report

In part (a), many candidates could not provide a logical sequence of steps to show that $$G$$ is cyclic. In particular, although they correctly quoted Lagrange’s theorem, they did not always consider all the orders of a, i.e., all the factors of 12, omitting in particular 1 as a factor. Some candidates did not state the second generator, in particular $${a^{ – 1}}$$. Very few candidates were successful in finding the required subgroup, although they were obviously familiar with setting up a Cayley table.

a.

b.

## Question

A group with the binary operation of multiplication modulo 15 is shown in the following Cayley table.

Find the values represented by each of the letters in the table.

[3]
a.

Find the order of each of the elements of the group.

[3]
b.

Write down the three sets that form subgroups of order 2.

[2]
c.

Find the three sets that form subgroups of order 4.

[4]
d.

## Markscheme

$$a = 1\;\;\;b = 8\;\;\;c = 4$$

$$d = 8\;\;\;e = 4\;\;\;f = 2$$

$$g = 4\;\;\;h = 2\;\;\;i = 1$$     A3

Note:     Award A3 for 9 correct answers, A2 for 6 or more, and A1 for 3 or more.

[3 marks]

a.

A3

Note:     Award A3 for 8 correct answers, A2 for 6 or more, and A1 for 4 or more.

[3 marks]

b.

$$\{ 1,{\text{ }}4\} ,{\text{ }}\{ 1,{\text{ }}11\} ,{\text{ }}\{ 1,{\text{ }}14\}$$     A1A1

Note:     Award A1 for 1 correct answer and A2 for all 3 (and no extras).

[2 marks]

c.

$$\{ 1,{\text{ }}2,{\text{ }}4,{\text{ }}8\} ,{\text{ }}\{ 1,{\text{ }}4,{\text{ }}7,{\text{ }}13\} ,$$     A1A1

$$\{ 1,{\text{ }}4,{\text{ }}11,{\text{ }}14\}$$     A2

[4 marks]

Total [12 marks]

d.

## Examiners report

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

a.

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

b.

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

c.

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

d.

## Question

Consider the set $${S_3} = \{ {\text{ }}p,{\text{ }}q,{\text{ }}r,{\text{ }}s,{\text{ }}t,{\text{ }}u\}$$ of permutations of the elements of the set $$\{ 1,{\text{ }}2,{\text{ }}3\}$$, defined by

$$p = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 1&2&3 \end{array}} \right),{\text{ }}q = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 1&3&2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 3&2&1 \end{array}} \right),{\text{ }}s = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&1&3 \end{array}} \right),{\text{ }}t = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&3&1 \end{array}} \right),{\text{ }}u = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 3&1&2 \end{array}} \right).$$

Let $$\circ$$ denote composition of permutations, so $$a \circ b$$ means $$b$$ followed by $$a$$. You may assume that $$({S_3},{\text{ }} \circ )$$ forms a group.

Complete the following Cayley table

[5 marks]

[4]
a.

(i)     State the inverse of each element.

(ii)     Determine the order of each element.

[6]
b.

Write down the subgroups containing

(i)     $$r$$,

(ii)     $$u$$.

[2]
c.

## Markscheme

(M1)A4

Note:     Award M1 for use of Latin square property and/or attempted multiplication, A1 for the first row or column, A1 for the squares of $$q$$, $$r$$ and $$s$$, then A2 for all correct.

a.

(i)     $${p^{ – 1}} = p,{\text{ }}{q^{ – 1}} = q,{\text{ }}{r^{ – 1}} = r,{\text{ }}{s^{ – 1}} = s$$     A1

$${t^{ – 1}} = u,{\text{ }}{u^{ – 1}} = t$$     A1

Note:     Allow FT from part (a) unless the working becomes simpler.

(ii)     using the table or direct multiplication     (M1)

the orders of $$\{ p,{\text{ }}q,{\text{ }}r,{\text{ }}s,{\text{ }}t,{\text{ }}u\}$$ are $$\{ 1,{\text{ }}2,{\text{ }}2,{\text{ }}2,{\text{ }}3,{\text{ }}3\}$$     A3

Note:     Award A1 for two, three or four correct, A2 for five correct.

[6 marks]

b.

(i)     $$\{ p,{\text{ }}r\} {\text{ }}\left( {{\text{and }}({S_3},{\text{ }} \circ )} \right)$$     A1

(ii)     $$\{ p,{\text{ }}u,{\text{ }}t\} {\text{ }}\left( {{\text{and }}({S_3},{\text{ }} \circ )} \right)$$     A1

Note:     Award A0A1 if the identity has been omitted.

Award A0 in (i) or (ii) if an extra incorrect “subgroup” has been included.

[2 marks]

Total [13 marks]

c.

## Examiners report

The majority of candidates were able to complete the Cayley table correctly. Unfortunately, many wasted time and space, laboriously working out the missing entries in the table – the identity is $$p$$ and the elements $$q$$, $$r$$ and $$s$$ are clearly of order two, so 14 entries can be filled in without any calculation. A few candidates thought $$t$$ and $$u$$ had order two.

a.

Generally well done. A few candidates were unaware of the definition of the order of an element.

b.

Often well done. A few candidates stated extra, and therefore incorrect subgroups.

c.

## Question

Consider the sets

$G = \left\{ {\frac{n}{{{6^i}}}|n \in \mathbb{Z},{\text{ }}i \in \mathbb{N}} \right\},{\text{ }}H = \left\{ {\frac{m}{{{3^j}}}|m \in \mathbb{Z},{\text{ }}j \in \mathbb{N}} \right\}.$

Show that $$(G,{\text{ }} + )$$ forms a group where $$+$$ denotes addition on $$\mathbb{Q}$$. Associativity may be assumed.

[5]
a.

Assuming that $$(H,{\text{ }} + )$$ forms a group, show that it is a proper subgroup of $$(G,{\text{ }} + )$$.

[4]
b.

The mapping $$\phi :G \to G$$ is given by $$\phi (g) = g + g$$, for $$g \in G$$.

Prove that $$\phi$$ is an isomorphism.

[7]
c.

## Markscheme

closure: $$\frac{{{n_1}}}{{{6^{{i_1}}}}} + \frac{{{n_2}}}{{{6^{{i_2}}}}} = \frac{{{6^{{i_2}}}{n_1} + {6^{{i_1}}}{n_2}}}{{{6^{{i_1} + {i_2}}}}} \in G$$     A1R1

Note:     Award A1 for RHS of equation. R1 is for the use of two different, but not necessarily most general elements, and the result $$\in G$$ or equivalent.

identity: $$0$$     A1

inverse: $$\frac{{ – n}}{{{6^i}}}$$     A1

since associativity is given, $$(G,{\text{ }} + )$$ forms a group     R1AG

Note:     The R1 is for considering closure, the identity, inverses and associativity.

[5 marks]

a.

it is required to show that $$H$$ is a proper subset of $$G$$     (M1)

let $$\frac{n}{{{3^i}}} \in H$$     M1

then $$\frac{n}{{{3^i}}} = \frac{{{2^i}n}}{{{6^i}}} \in G$$ hence $$H$$ is a subgroup of $$G$$     A1

$$H \ne G$$ since $$\frac{1}{6} \in G$$ but $$\frac{1}{6} \notin H$$     A1

Note:     The final A1 is only dependent on the first M1.

hence, $$H$$ is a proper subgroup of $$G$$     AG

[4 marks]

b.

consider $$\phi ({g_1} + {g_2}) = ({g_1} + {g_2}) + ({g_1} + {g_2})$$     M1

$$= ({g_1} + {g_1}) + ({g_2} + {g_2}) = \phi ({g_1}) + \phi ({g_2})$$     A1

(hence $$\phi$$ is a homomorphism)

injectivity: let $$\phi ({g_1}) = \phi ({g_2})$$     M1

working within $$\mathbb{Q}$$ we have $$2{g_1} = 2{g_2} \Rightarrow {g_1} = {g_2}$$     A1

surjectivity: considering even and odd numerators     M1

$$\phi \left( {\frac{n}{{{6^i}}}} \right) = \frac{{2n}}{{{6^i}}}$$ and $$\phi \left( {\frac{{3(2n + 1)}}{{{6^{i + 1}}}}} \right) = \frac{{2n + 1}}{{{6^i}}}$$     A1A1

hence $$\phi$$ is an isomorphism     AG

[7 marks]

Total [16 marks]

c.

## Examiners report

This part was generally well done. Where marks were lost, it was usually because a candidate failed to choose two different elements in the proof of closure.

a.

Only a few candidates realised that they did not have to prove that $$H$$ is a group – that was stated in the question. Some candidates tried to invoke Lagrange’s theorem, even though $$G$$ is an infinite group.

b.

Many candidates showed that the mapping is injective. Most attempts at proving surjectivity were unconvincing. Those candidates who attempted to establish the homomorphism property sometimes failed to use two different elements.

c.

## Question

The set of all permutations of the elements $$1,{\text{ }}2,{\text{ }} \ldots 10$$ is denoted by $$H$$ and the binary operation $$\circ$$ represents the composition of permutations.

The permutation $$p = (1{\text{ }}2{\text{ }}3{\text{ }}4{\text{ }}5{\text{ }}6)(7{\text{ }}8{\text{ }}9{\text{ }}10)$$ generates the subgroup $$\{ G,{\text{ }} \circ \}$$ of the group $$\{ H,{\text{ }} \circ \}$$.

Find the order of $$\{ G,{\text{ }} \circ \}$$.

[2]
a.

State the identity element in $$\{ G,{\text{ }} \circ \}$$.

[1]
b.

Find

(i)     $$p \circ p$$;

(ii)     the inverse of $$p \circ p$$.

[4]
c.

(i)     Find the maximum possible order of an element in $$\{ H,{\text{ }} \circ \}$$.

(ii)     Give an example of an element with this order.

[3]
d.

## Markscheme

the order of $$(G,{\text{ }} \circ )$$ is $${\text{lcm}}(6,{\text{ }}4)$$     (M1)

$$= 12$$     A1

[2 marks]

a.

$$\left( 1 \right){\rm{ }}\left( 2 \right){\rm{ }}\left( 3 \right){\rm{ }}\left( 4 \right){\rm{ }}\left( 5 \right){\rm{ }}\left( 6 \right){\rm{ }}\left( 7 \right){\rm{ }}\left( 8 \right){\rm{ }}\left( 9 \right){\rm{ }}\left( {10} \right)$$     A1

Note:     Accept ( ) or a word description.

[1 mark]

b.

(i)     $$p \circ p = (1{\text{ }}3{\text{ }}5)(2{\text{ }}4{\text{ }}6)(7{\text{ }}9)(810)$$     (M1)A1

(ii)     its inverse $$= (1{\text{ }}5{\text{ }}3)(2{\text{ }}6{\text{ }}4)(7{\text{ }}9)(810)$$     A1A1

Note:     Award A1 for cycles of 2, A1 for cycles of 3.

[4 marks]

c.

(i)     considering LCM of length of cycles with length $$2$$, $$3$$ and $$5$$     (M1)

$$30$$     A1

(ii)     eg$$\;\;\;(1{\text{ }}2)(3{\text{ }}4{\text{ }}5)(6{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}10)$$     A1

Note:     allow FT as long as the length of cycles adds to $$10$$ and their LCM is consistent with answer to part (i).

Note: Accept alternative notation for each part

[3 marks]

Total [10 marks]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

## Question

The following Cayley table for the binary operation multiplication modulo 9, denoted by $$*$$, is defined on the set $$S = \{ 1,{\text{ }}2,{\text{ }}4,{\text{ }}5,{\text{ }}7,{\text{ }}8\}$$.

Copy and complete the table.

[3]
a.

Show that $$\{ S,{\text{ }} * \}$$ is an Abelian group.

[5]
b.

Determine the orders of all the elements of $$\{ S,{\text{ }} * \}$$.

[3]
c.

(i)     Find the two proper subgroups of $$\{ S,{\text{ }} * \}$$.

(ii)     Find the coset of each of these subgroups with respect to the element 5.

[4]
d.

Solve the equation $$2 * x * 4 * x * 4 = 2$$.

[4]
e.

## Markscheme

A3

Note:     Award A3 for correct table, A2 for one or two errors, A1 for three or four errors and A0 otherwise.

[3 marks]

a.

the table contains only elements of $$S$$, showing closure     R1

the identity is 1     A1

every element has an inverse since 1 appears in every row and column, or a complete list of elements and their correct inverses     A1

multiplication of numbers is associative     A1

the four axioms are satisfied therefore $$\{ S,{\text{ }} * \}$$ is a group

the group is Abelian because the table is symmetric (about the leading diagonal)     A1

[5 marks]

b.

A3

Note:     Award A3 for all correct values, A2 for 5 correct, A1 for 4 correct and A0 otherwise.

[3 marks]

c.

(i)     the subgroups are $$\{ 1,{\text{ }}8\}$$; $$\{ 1,{\text{ }}4,{\text{ }}7\}$$     A1A1

(ii)     the cosets are $$\{ 4,{\text{ }}5\}$$; $$\{ 2,{\text{ }}5,{\text{ }}8\}$$     A1A1

[4 marks]

d.

METHOD 1

use of algebraic manipulations     M1

and at least one result from the table, used correctly     A1

$$x = 2$$    A1

$$x = 7$$    A1

METHOD 2

testing at least one value in the equation     M1

obtain $$x = 2$$     A1

obtain $$x = 7$$     A1

explicit rejection of all other values     A1

[4 marks]

e.

## Examiners report

The majority of candidates were able to complete the Cayley table correctly.

a.

Generally well done. However, it is not good enough for a candidate to say something along the lines of ‘the operation is closed or that inverses exist by looking at the Cayley table’. A few candidates thought they only had to prove commutativity.

b.

Often well done. A few candidates stated extra, and therefore incorrect subgroups.

c.

[N/A]

d.

The majority found only one solution, usually the obvious $$x = 2$$, but sometimes only the less obvious $$x = 7$$.

e.

## Question

The group $$\{ G,{\text{ }} * \}$$ is defined on the set $$G$$ with binary operation $$*$$. $$H$$ is a subset of $$G$$ defined by $$H = \{ x:{\text{ }}x \in G,{\text{ }}a * x * {a^{ – 1}} = x{\text{ for all }}a \in G\}$$. Prove that $$\{ H,{\text{ }} * \}$$ is a subgroup of $$\{ G,{\text{ }} * \}$$.

## Markscheme

associativity: This follows from associativity in $$\{ G,{\text{ }} * \}$$     R1

the identity $$e \in H$$ since $$a * e * {a^{ – 1}} = a * {a^{ – 1}} = e$$ (for all $$a \in G$$)     R1

Note:     Condone the use of the commutativity of e if that is involved in an alternative simplification of the LHS.

closure: Let $$x,{\text{ }}y \in H$$ so that $$a * x * {a^{ – 1}} = x$$ and $$a * y * {a^{ – 1}} = y$$ for all $$a \in G$$     (M1)

multiplying, $$x * y = a * x * {a^{ – 1}} * a * y * {a^{ – 1}}$$ (for all $$a \in G$$)     A1

$$= a * x * y * {a^{ – 1}}$$    A1

therefore $$x * y \in H$$ (proving closure)     R1

inverse: Let $$x \in H$$ so that $$a * x * {a^{ – 1}} = x$$ (for all $$a \in G$$)

$${x^{ – 1}} = {(a * x * {a^{ – 1}})^{ – 1}}$$    M1

$$= a * {x^{ – 1}} * {a^{ – 1}}$$    A1

therefore $${x^{ – 1}} \in H$$     R1

hence $$\{ H,{\text{ }} * \}$$ is a subgroup of $$\{ G,{\text{ }} * \}$$     AG

Note:     Accuracy marks cannot be awarded if commutativity is assumed for general elements of $$G$$.

[9 marks]

## Examiners report

This is an abstract question, clearly defined on a subset. Far too many candidates almost immediately deduced, erroneously, that the full group was Abelian. Almost no marks were then available.

## Question

Consider the group $$\{ G,{\text{ }}{ \times _{18}}\}$$ defined on the set $$\{ 1,{\text{ }}5,{\text{ }}7,{\text{ }}11,{\text{ }}13,{\text{ }}17\}$$ where $${ \times _{18}}$$ denotes multiplication modulo 18. The group $$\{ G,{\text{ }}{ \times _{18}}\}$$ is shown in the following Cayley table.

The subgroup of $$\{ G,{\text{ }}{ \times _{18}}\}$$ of order two is denoted by $$\{ K,{\text{ }}{ \times _{18}}\}$$.

Find the order of elements 5, 7 and 17 in $$\{ G,{\text{ }}{ \times _{18}}\}$$.

[4]
a.i.

State whether or not $$\{ G,{\text{ }}{ \times _{18}}\}$$ is cyclic, justifying your answer.

[2]
a.ii.

Write down the elements in set $$K$$.

[1]
b.

Find the left cosets of $$K$$ in $$\{ G,{\text{ }}{ \times _{18}}\}$$.

[4]
c.

## Markscheme

considering powers of elements     (M1)

5 has order 6     A1

7 has order 3     A1

17 has order 2     A1

[4 marks]

a.i.

$$G$$ is cyclic     A1

because there is an element (are elements) of order 6     R1

Note:     Accept “there is a generator”; allow A1R0.

[3 marks]

a.ii.

$$\{ 1,{\text{ }}17\}$$     A1

[1 mark]

b.

multiplying $$\{ 1,{\text{ }}17\}$$ by each element of $$G$$     (M1)

$$\{ 1,{\text{ }}17\} ,{\text{ }}\{ 5,{\text{ }}13\} ,{\text{ }}\{ 7,{\text{ }}11\}$$     A1A1A1

[4 marks]

c.

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

[N/A]

c.

## Question

Let $$\{ G,{\text{ }} * \}$$ be a finite group and let H be a non-empty subset of G . Prove that $$\{ H,{\text{ }} * \}$$ is a group if H is closed under $$*$$.

## Markscheme

the associativity property carries over from G     R1

closure is given     R1

let $$h \in H$$ and let n denote the order of h, (this is finite because G is finite)     M1

it follows that $${h^n} = e$$, the identity element     R1

and since H is closed, $$e \in H$$     R1

since $$h * {h^{n – 1}} = e$$     M1

it follows that $${h^{n – 1}}$$ is the inverse, $${h^{ – 1}}$$, of h     R1

and since H is closed, $${h^{ – 1}} \in H$$ so each element of H has an inverse element     R1

the four requirements for H to be a group are therefore satisfied     AG

[8 marks]

[N/A]