IB DP Maths Topic 8.11 Lagrange’s theorem HL Paper 3

 

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Question

(a)     Write down why the table below is a Latin square.

\[\begin{gathered}
  \begin{array}{*{20}{c}}
  {}&d&e&b&a&c
\end{array} \\
  \begin{array}{*{20}{c}}
  d \\
  e \\
  b \\
  a \\
  c
\end{array}\left[ {\begin{array}{*{20}{c}}
  c&d&e&b&a \\
  d&e&b&a&c \\
  a&b&d&c&e \\
  b&a&c&e&d \\
  e&c&a&d&b
\end{array}} \right] \\
\end{gathered} \]

(b)     Use Lagrange’s theorem to show that the table is not a group table.

Answer/Explanation

Markscheme

(a)     Each row and column contains all the elements of the set.     A1A1

[2 marks]

 

(b)     There are 5 elements therefore any subgroup must be of an order that is a factor of 5     R2

But there is a subgroup \(\begin{gathered}
  \begin{array}{*{20}{c}}
  {}&e&a
\end{array} \\
  \begin{array}{*{20}{c}}
  e \\
  a
\end{array}\left( {\begin{array}{*{20}{c}}
  e&a \\
  a&e
\end{array}} \right) \\
\end{gathered} \) of order 2 so the table is not a group table     R2

Note: Award R0R2 for “a is an element of order 2 which does not divide the order of the group”.

 

[4 marks]

Total [6 marks]

Examiners report

Part (a) presented no problem but finding the order two subgroups (Lagrange’s theorem was often quoted correctly) was beyond some candidates. Possibly presenting the set in non-alphabetical order was the problem.

Question

\(\{ G,{\text{ }} * \} \) is a group with identity element \(e\). Let \(a,{\text{ }}b \in G\).

State Lagrange’s theorem.

[2]
a.

Verify that the inverse of \(a * {b^{ – 1}}\) is equal to \(b * {a^{ – 1}}\).

[3]
b.

Let \(\{ H,{\rm{ }} * {\rm{\} }}\) be a subgroup of \(\{ G,{\rm{ }} * {\rm{\} }}\). Let \(R\) be a relation defined on \(G\) by

\[aRb \Leftrightarrow a * {b^{ – 1}} \in H.\]

Prove that \(R\) is an equivalence relation, indicating clearly whenever you are using one of the four properties required of a group.

[8]
c.

Let \(\{ H,{\rm{ }} * {\rm{\} }}\) be a subgroup of \(\{ G,{\rm{ }} * {\rm{\} }}\) .Let \(R\) be a relation defined on \(G\) by

\[aRb \Leftrightarrow a * {b^{ – 1}} \in H.\]

Show that \(aRb \Leftrightarrow a \in Hb\), where \(Hb\) is the right coset of \(H\) containing \(b\).

[3]
d.

Let \(\{ H,{\rm{ }} * {\rm{\} }}\) be a subgroup of \(\{ G,{\rm{ }} * {\rm{\} }}\) .Let \(R\) be a relation defined on \(G\) by

\[aRb \Leftrightarrow a * {b^{ – 1}} \in H.\]

It is given that the number of elements in any right coset of \(H\) is equal to the order of \(H\).

Explain how this fact together with parts (c) and (d) prove Lagrange’s theorem.

[3]
e.
Answer/Explanation

Markscheme

in a finite group the order of any subgroup (exactly) divides the order of the group     A1A1

[2 marks]

a.

METHOD 1

\((a * {b^{ – 1}}) * (b * {a^{ – 1}}) = a * {b^{ – 1}} * b * {a^{ – 1}} = a * e * {a^{ – 1}} = a * {a^{ – 1}} = e\)     M1A1A1

Note:     M1 for multiplying, A1 for at least one of the next 3 expressions,

A1 for \(e\).

Allow \((b * {a^{ – 1}}) * (a * {b^{ – 1}}) = b * {a^{ – 1}} * a * {b^{ – 1}} = b * e * {b^{ – 1}} = b * {b^{ – 1}} = e\).

METHOD 2

\({(a * {b^{ – 1}})^{ – 1}} = {({b^{ – 1}})^{ – 1}} * {a^{ – 1}}\)     M1A1

\( = b * {a^{ – 1}}\)A1

[3 marks]

b.

\(a * {a^{ – 1}} = e \in H\;\;\;\)(as \(H\) is a subgroup)     M1

so \(aRa\) and hence \(R\) is reflexive

\(aRb \Leftrightarrow a * {b^{ – 1}} \in H\). \(H\) is a subgroup so every element has an inverse in \(H\) so

\({(a * {b^{ – 1}})^{ – 1}} \in H\)     R1

\( \Leftrightarrow b * {a^{ – 1}} \in H \Leftrightarrow bRa\)     M1

so \(R\) is symmetric

\(aRb,{\text{ }}bRc \Leftrightarrow a * {b^{ – 1}} \in H,{\text{ }}b * {c^{ – 1}} \in H\)     M1

as \(H\) is closed \((a * {b^{ – 1}}) * {\text{(}}b * {c^{ – 1}}) \in H\)     R1

and using associativity     R1

\((a * {b^{ – 1}}) * {\text{(}}b * {c^{ – 1}}) = a * ({b^{ – 1}} * b) * {c^{ – 1}} = a * {c^{ – 1}} \in H \Leftrightarrow aRc\)     A1

therefore \(R\) is transitive

\(R\) is reflexive, symmetric and transitive

Note:     Can be said separately at the end of each part.

hence it is an equivalence relation     AG

[8 marks]

c.

\(aRb \Leftrightarrow a * {b^{ – 1}} \in H \Leftrightarrow a * {b^{ – 1}} = h \in H\)     A1

\( \Leftrightarrow a = h * b \Leftrightarrow a \in Hb\)     M1R1

[3 marks]

d.

(d) implies that the right cosets of \(H\) are equal to the equivalence classes of the relation in (c)     R1

hence the cosets partition \(G\)     R1

all the cosets are of the same size as the subgroup \(H\) so the order of \(G\) must be a multiple of \(\left| H \right|\)     R1

[3 marks]

Total [19 marks]

e.

Examiners report

Many students obtained just half marks in (a) for not stating the requirement of the order to be finite.

a.

Part (b) should have been more straightforward than many found.

b.

In part (c) it was evident that most candidates knew what to do, but being a more difficult question fell down on a lack of rigour. Nonetheless, many candidates obtained full or partial marks on this question part.

c.

Part (d) enabled many candidates to obtain, at least partial marks, but there were few students with the insight to be able to answer part (e) satisfactorily.

d.

Part (d) enabled many candidates to obtain, at least partial marks, but there were few students with the insight to be able to answer part (e) satisfactorily.

e.

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.
Answer/Explanation

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 \(\{ G,{\text{ }} * \} \).

Answer/Explanation

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.
Answer/Explanation

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.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

[N/A]

c.

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