IB DP Maths Topic 8.12 Isomorphism of groups. HL Paper 3

Question

(a)     Find the six roots of the equation $${z^6} – 1 = 0$$ , giving your answers in the form $$r\,{\text{cis}}\,\theta {\text{, }}r \in {\mathbb{R}^ + }{\text{, }}0 \leqslant \theta < 2\pi$$ .

(b)     (i)     Show that these six roots form a group G under multiplication of complex numbers.

(ii)     Show that G is cyclic and find all the generators.

(iii)     Give an example of another group that is isomorphic to G, stating clearly the corresponding elements in the two groups.

Markscheme

(a)     $${z^6} = 1 = {\text{cis}}\,2n\pi$$     (M1)

The six roots are

$${\text{cis}}\,0(1),{\text{ cis}}\frac{\pi }{3},{\text{ cis}}\frac{{2\pi }}{3},{\text{ cis}}\,\pi ( – 1),{\text{ cis}}\frac{{4\pi }}{3},{\text{ cis}}\frac{{5\pi }}{3}$$     A3

Note: Award A2 for 4 or 5 correct roots, A1 for 2 or 3 correct roots.

[4 marks]

(b)     (i)     Closure: Consider any two roots $${\text{cis}}\frac{{m\pi }}{3},{\text{ cis}}\frac{{n\pi }}{3}$$.     M1

$${\text{cis}}\frac{{m\pi }}{3} \times {\text{cis}}\frac{{n\pi }}{3} = {\text{cis}}\,(m + n){\text{(mod6)}}\frac{\pi }{3} \in G$$     A1

Note: Award M1A1 for a correct Cayley table showing closure.

Identity: The identity is 1.     A1

Inverse: The inverse of $${\text{cis}}\frac{{m\pi }}{3}{\text{ is cis}}\frac{{(6 – m)\pi }}{3} \in G$$ .     A2

Associative: This follows from the associativity of multiplication.     R1

The 4 group axioms are satisfied.     R1

(ii)     Successive powers of $${\text{cis}}\frac{\pi }{3}\left( {{\text{or cis}}\frac{{5\pi }}{3}} \right)$$

generate the group which is therefore cyclic.     R2

The (only) other generator is $${\text{cis}}\frac{{5\pi }}{3}\left( {{\text{or cis}}\frac{\pi }{3}} \right)$$ .     A1

(iii)     The group of the integers 0, 1, 2, 3, 4, 5 under addition modulo 6.     R2

The correspondence is

$$m \to {\text{cis}}\frac{{m\pi }}{3}$$     R1

Note: Accept any other cyclic group of order 6.

[13 marks]

Total [17 marks]

Examiners report

This question was reasonably well answered by many candidates, although in (b)(iii), some candidates were unable to give another group isomorphic to G.

Question

(a)     Consider the set A = {1, 3, 5, 7} under the binary operation $$*$$, where $$*$$ denotes multiplication modulo 8.

(i)     Write down the Cayley table for $$\{ A,{\text{ }} * \}$$.

(ii)     Show that $$\{ A,{\text{ }} * \}$$ is a group.

(iii)     Find all solutions to the equation $$3 * x * 7 = y$$. Give your answers in the form $$(x,{\text{ }}y)$$.

(b)     Now consider the set B = {1, 3, 5, 7, 9} under the binary operation $$\otimes$$, where $$\otimes$$ denotes multiplication modulo 10. Show that $$\{ B,{\text{ }} \otimes \}$$ is not a group.

(c)     Another set C can be formed by removing an element from B so that $$\{ C,{\text{ }} \otimes \}$$ is a group.

(i)     State which element has to be removed.

(ii)     Determine whether or not $$\{ A,{\text{ }} * \}$$ and $$\{ C,{\text{ }} \otimes \}$$ are isomorphic.

Markscheme

(a)     (i)          A3

Note: Award A2 for 15 correct, A1 for 14 correct and A0 otherwise.

(ii)     it is a group because:

the table shows closure     A1

multiplication is associative     A1

it possesses an identity 1     A1

justifying that every element has an inverse e.g. all self-inverse     A1

(iii)     (since $$*$$ is commutative, $$5 * x = y$$)

so solutions are (1, 5), (3, 7), (5, 1), (7, 3)     A2

Notes: Award A1 for 3 correct and A0 otherwise.

Do not penalize extra incorrect solutions.

[9 marks]

(b)

Note: It is not necessary to see the Cayley table.

a valid reason     R2

e.g. from the Cayley table the 5 row does not give a Latin square, or 5 does not have an inverse, so it cannot be a group

[2 marks]

(c)     (i)     remove the 5     A1

(ii)     they are not isomorphic because all elements in A are self-inverse this is not the case in C, (e.g. $$3 \otimes 3 = 9 \ne 1$$)     R2

Note: Accept any valid reason.

[3 marks]

Total [14 marks]

Examiners report

Candidates are generally confident when dealing with a specific group and that was the situation again this year. Some candidates lost marks in (a)(ii) by not giving an adequate explanation for the truth of some of the group axioms, eg some wrote ‘every element has an inverse’. Since the question told the candidates that $$\{ A,{\text{ }} * )$$ was a group, this had to be the case and the candidates were expected to justify their statement by noting that every element was self-inverse. Solutions to (c)(ii) were reasonably good in general, certainly better than solutions to questions involving isomorphisms set in previous years.

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

Given that p , q and r are elements of a group, prove the left-cancellation rule, i.e. $$pq = pr \Rightarrow q = r$$ .

Your solution should indicate which group axiom is used at each stage of the proof.

[4]
a.

Consider the group G , of order 4, which has distinct elements a , b and c and the identity element e .

(i)     Giving a reason in each case, explain why ab cannot equal a or b .

(ii)     Given that c is self inverse, determine the two possible Cayley tables for G .

(iii)     Determine which one of the groups defined by your two Cayley tables is isomorphic to the group defined by the set {1, −1, i, −i} under multiplication of complex numbers. Your solution should include a correspondence between a, b, c, e and 1, −1, i, −i .

[10]
b.

Markscheme

$$pq = pr$$

$${p^{ – 1}}(pq) = {p^{ – 1}}(pr)$$ , every element has an inverse     A1

$$({p^{ – 1}}p)q = ({p^{ – 1}}p)r$$ , Associativity     A1

Note: Brackets in lines 2 and 3 must be seen.

$$eq = er$$, $${p^{ – 1}}p = e$$, the identity     A1

$$q = r$$, $$ea = a$$ for all elements a of the group     A1

[4 marks]

a.

(i)     let ab = a so b = e be which is a contradiction     R1

let ab = b so a = e which is a contradiction     R1

therefore ab cannot equal either a or b     AG

(ii)     the two possible Cayley tables are

table 1

A2

table 2

A2

(iii)     the group defined by table 1 is isomorphic to the given group     R1

because

EITHER

both contain one self-inverse element (other than the identity)     R1

OR

both contain an inverse pair     R1

OR

both are cyclic     R1

THEN

the correspondence is $$e \to 1$$, $$c \to – 1$$, $$a \to i$$, $$b \to – i$$

(or vice versa for the last two)     A2

Note: Award the final A2 only if the correct group table has been identified.

[10 marks]

b.

Examiners report

Solutions to (a) were often poor with inadequate explanations often seen. It was not uncommon to see $$pq = pr$$

$${p^{ – 1}}pq = {p^{ – 1}}pr$$

$$q = r$$

without any mention of associativity. Many candidates understood what was required in (b)(i), but solutions to (b)(ii) were often poor with the tables containing elements such as ab and bc without simplification. In (b)(iii), candidates were expected to determine the isomorphism by noting that the group defined by {1, –1, i, –i} under multiplication is cyclic or that –1 is the only self-inverse element apart from the identity, without necessarily writing down the Cayley table in full which many candidates did. Many candidates just stated that there was a bijection between the two groups without giving any justification for this.

a.

Solutions to (a) were often poor with inadequate explanations often seen. It was not uncommon to see $$pq = pr$$

$${p^{ – 1}}pq = {p^{ – 1}}pr$$

$$q = r$$

without any mention of associativity. Many candidates understood what was required in (b)(i), but solutions to (b)(ii) were often poor with the tables containing elements such as ab and bc without simplification. In (b)(iii), candidates were expected to determine the isomorphism by noting that the group defined by {1, –1, i, –i} under multiplication is cyclic or that –1 is the only self-inverse element apart from the identity, without necessarily writing down the Cayley table in full which many candidates did. Many candidates just stated that there was a bijection between the two groups without giving any justification for this.

b.

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

Let $$A = \left\{ {a,{\text{ }}b} \right\}$$.

Let the set of all these subsets be denoted by $$P(A)$$ . The binary operation symmetric difference, $$\Delta$$ , is defined on $$P(A)$$ by $$X\Delta Y = (X\backslash Y) \cup (Y\backslash X)$$ where $$X$$ , $$Y \in P(A)$$.

Let $${\mathbb{Z}_4} = \left\{ {0,{\text{ }}1,{\text{ }}2,{\text{ }}3} \right\}$$ and $${ + _4}$$ denote addition modulo $$4$$.

Let $$S$$ be any non-empty set. Let $$P(S)$$ be the set of all subsets of $$S$$ . For the following parts, you are allowed to assume that $$\Delta$$, $$\cup$$ and $$\cap$$ are associative.

Write down all four subsets of A .

[1]
a.

Construct the Cayley table for $$P(A)$$ under $$\Delta$$ .

[3]
b.

Prove that $$\left\{ {P(A),{\text{ }}\Delta } \right\}$$ is a group. You are allowed to assume that $$\Delta$$ is associative.

[3]
c.

Is $$\{ P(A){\text{, }}\Delta \}$$ isomorphic to $$\{ {\mathbb{Z}_4},{\text{ }}{ + _4}\}$$ ? Justify your answer.

[2]
d.

(i)     State the identity element for $$\{ P(S){\text{, }}\Delta \}$$.

(ii)     Write down $${X^{ – 1}}$$ for $$X \in P(S)$$ .

(iii)     Hence prove that $$\{ P(S){\text{, }}\Delta \}$$ is a group.

[4]
e.

Explain why $$\{ P(S){\text{, }} \cup \}$$ is not a group.

[1]
f.

Explain why $$\{ P(S){\text{, }} \cap \}$$ is not a group.

[1]
g.

Markscheme

$$\emptyset {\text{, \{ a\} , \{ b\} , \{ a, b\} }}$$     A1

[1 mark]

a.

A3

Note: Award A2 for one error, A1 for two errors, A0 for more than two errors.

[3 marks]

b.

closure is seen from the table above     A1

$$\emptyset$$ is the identity     A1

each element is self-inverse     A1

Note: Showing each element has an inverse is sufficient.

associativity is assumed so we have a group     AG

[3 marks]

c.

not isomorphic as in the above group all elements are self-inverse whereas in $$({\mathbb{Z}_4},{\text{ }}{ + _4})$$ there is an element of order 4 (e.g. 1)     R2

[2 marks]

d.

(i)     $$\emptyset$$ is the identity     A1

(ii)     $${X^{ – 1}} = X$$     A1

(iii)     if X and Y are subsets of S then $$X\Delta Y$$ (the set of elements that belong to X or Y but not both) is also a subset of S, hence closure is proved     R1

$$\{ P(S){\text{, }}\Delta \}$$ is a group because it is closed, has an identity, all elements have inverses (and $$\Delta$$ is associative)     R1AG

[4 marks]

e.

not a group because although the identity is $$\emptyset {\text{, if }}X \ne \emptyset$$ it is impossible to find a set Y such that $$X \cup Y = \emptyset$$, so there are elements without an inverse     R1AG

[1 mark]

f.

not a group because although the identity is S, if $$X \ne S$$ is impossible to find a set Y such that $$X \cap Y = S$$, so there are elements without an inverse     R1AG

[1 mark]

g.

Examiners report

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

a.

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

b.

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

c.

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

d.

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

e.

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

f.

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

g.

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 group $$\{ G,{\text{ }} * \}$$ is Abelian and the bijection $$f:{\text{ }}G \to G$$ is defined by $$f(x) = {x^{ – 1}},{\text{ }}x \in G$$.

Show that $$f$$ is an isomorphism.

Markscheme

we need to show that $$f(a * b) = f(a) * f(b)$$     R1

Note:     This R1 may be awarded at any stage.

let $$a,{\text{ }}b \in G$$     (M1)

consider $$f(a) * f(b)$$     M1

$$= {a^{ – 1}} * {b^{ – 1}}$$    A1

consider $$f(a * b) = {(a * b)^{ – 1}}$$     M1

$$= {b^{ – 1}} * {a^{ – 1}}$$    A1

$$= {a^{ – 1}} * {b^{ – 1}}$$ since $$G$$ is Abelian     R1

hence $$f$$ is an isomorphism     AG

[7 marks]

Examiners report

A surprising number of candidates wasted time and unrewarded effort showing that the mapping $$f$$, stated to be a bijection in the question, actually was a bijection. Many candidates failed to get full marks by not properly using the fact that the group was stated to be Abelian. There were also candidates who drew the graph of $$y = \frac{1}{x}$$ or otherwise assumed that the inverse of $$x$$ was its reciprocal – this is unacceptable in the context of an abstract group question.

Question

The set of all permutations of the list of the integers 1, 2, 3  4 is a group, S4, under the operation of function composition.

In the group S4 let $${p_1} = \left( \begin{gathered} \begin{array}{*{20}{c}} 1&2&3&4 \end{array} \hfill \\ \begin{array}{*{20}{c}} 2&3&1&4 \end{array} \hfill \\ \end{gathered} \right)$$ and $${p_2} = \left( \begin{gathered} \begin{array}{*{20}{c}} 1&2&3&4 \end{array} \hfill \\ \begin{array}{*{20}{c}} 2&1&3&4 \end{array} \hfill \\ \end{gathered} \right)$$.

Determine the order of S4.

[2]
a.

Find the proper subgroup H of order 6 containing $${p_1}$$, $${p_2}$$ and their compositions. Express each element of H in cycle form.

[5]
b.

Let $$f{\text{:}}\,{S_4} \to {S_4}$$ be defined by $$f\left( p \right) = p \circ p$$ for $$p \in {S_4}$$.

Using $${p_1}$$ and $${p_2}$$, explain why $$f$$ is not a homomorphism.

[5]
c.

Markscheme

number of possible permutations is 4 × 3 × 2 × 1       (M1)

= 24(= 4!)      A1

[2 marks]

a.

attempting to find one of $${p_1} \circ {p_1}$$, $${p_1} \circ {p_2}$$ or $${p_2} \circ {p_1}$$     M1

$${p_1} \circ {p_1} = \left( {132} \right)$$ or equivalent (eg, $${p_1}^{ – 1} = \left( {132} \right)$$)    A1

$${p_1} \circ {p_2} = \left( {13} \right)$$ or equivalent (eg, $${p_2} \circ {p_1} \circ {p_1} = \left( {13} \right)$$)    A1

$${p_2} \circ {p_1} = \left( {23} \right)$$ or equivalent (eg, $${p_1} \circ {p_1} \circ {p_2} = \left( {23} \right)$$)    A1

Note: Award A1A0A0 for one correct permutation in any form; A1A1A0 for two correct permutations in any form.

$$e = \left( 1 \right)$$, $${p_1} = \left( {123} \right)$$ and $${p_2} = \left( {12} \right)$$     A1

Note: Condone omission of identity in cycle form as long as it is clear it is considered one of the elements of H.

[5 marks]

b.

METHOD 1

if $$f$$ is a homomorphism $$f\left( {{p_1} \circ {p_2}} \right) = f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$

attempting to express one of $$f\left( {{p_1} \circ {p_2}} \right)$$ or $$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$ in terms of $${p_1}$$ and $${p_2}$$      M1

$$f\left( {{p_1} \circ {p_2}} \right) = {p_1} \circ {p_2} \circ {p_1} \circ {p_2}$$     A1

$$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right) = {p_1} \circ {p_1} \circ {p_2} \circ {p_2}$$     A1

$$\Rightarrow {p_2} \circ {p_1} = {p_1} \circ {p_2}$$     A1

but $${p_1} \circ {p_2} \ne {p_2} \circ {p_1}$$     R1

so $$f$$ is not a homomorphism     AG

Note: Award R1 only if M1 is awarded.

Note: Award marks only if $${p_1}$$ and $${p_2}$$ are used; cycle form is not required.

METHOD 2

if $$f$$ is a homomorphism $$f\left( {{p_1} \circ {p_2}} \right) = f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$

attempting to find one of $$f\left( {{p_1} \circ {p_2}} \right)$$ or $$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$      M1

$$f\left( {{p_1} \circ {p_2}} \right) = e$$     A1

$$f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right) = \left( {132} \right)$$     (M1)A1

so $$f\left( {{p_1} \circ {p_2}} \right) \ne f\left( {{p_1}} \right) \circ f\left( {{p_2}} \right)$$     R1

so $$f$$ is not a homomorphism     AG

Note: Award R1 only if M1 is awarded.

Note: Award marks only if $${p_1}$$ and $${p_2}$$ are used; cycle form is not required.

[5 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

Question

The group $$\{ G,{\text{ }}{ \times _7}\}$$ is defined on the set {1, 2, 3, 4, 5, 6} where $${ \times _7}$$ denotes multiplication modulo 7.

(i)     Write down the Cayley table for $$\{ G,{\text{ }}{ \times _7}\}$$ .

(ii)     Determine whether or not $$\{ G,{\text{ }}{ \times _7}\}$$ is cyclic.

(iii)     Find the subgroup of G of order 3, denoting it by H .

(iv)     Identify the element of order 2 in G and find its coset with respect to H .

[10]
a.

The group $$\{ K,{\text{ }} \circ \}$$ is defined on the six permutations of the integers 1, 2, 3 and $$\circ$$ denotes composition of permutations.

(i)     Show that $$\{ K,{\text{ }} \circ \}$$ is non-Abelian.

(ii)     Giving a reason, state whether or not $$\{ G,{\text{ }}{ \times _7}\}$$ and $$\{ K,{\text{ }} \circ \}$$ are isomorphic.

[6]
b.

Markscheme

(i)     the Cayley table is

A3

Note: Deduct 1 mark for each error up to a maximum of 3.

(ii)     by considering powers of elements,     (M1)

it follows that 3 (or 5) is of order 6     A1

so the group is cyclic     A1

(iii)     we see that 2 and 4 are of order 3 so the subgroup of order 3 is {1, 2, 4}     M1A1

(iv)     the element of order 2 is 6     A1

the coset is {3, 5, 6}     A1

[10 marks]

a.

(i)     consider for example

$$\left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&1&3 \end{array}} \right) \circ \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&3&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 1&3&2 \end{array}} \right)$$     M1A1

$$\left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&3&1 \end{array}} \right) \circ \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&1&3 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 3&2&1 \end{array}} \right)$$     M1A1

Note: Award M1A1M1A0 if both compositions are done in the wrong order.

Note: Award M1A1M0A0 if the two compositions give the same result, if no further attempt is made to find two permutations which are not commutative.

these are different so the group is not Abelian     R1AG

(ii)     they are not isomorphic because $$\{ G,{\text{ }}{ \times _7}\}$$ is Abelian and $$\{ K,{\text{ }} \circ \}$$ is not     R1

[6 marks]

b.

[N/A]

a.

[N/A]

b.