IB DP Maths Topic 8.10 The order of a combination of cycles HL Paper 3

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 \({p_2} = \left( \begin{gathered}
\begin{array}{*{20}{c}}
1&2&3&4
\end{array} \hfill \\
\begin{array}{*{20}{c}}
2&1&3&4
\end{array} \hfill \\
\end{gathered} \right)\).

a.Determine the order of S4.[2]

b.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]

c.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]

▶️Answer/Explanation

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]

 

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 \(\{ H,{\text{ }} \circ \} \).

a.Find the order of \(\{ G,{\text{ }} \circ \} \).[2]

b.State the identity element in \(\{ G,{\text{ }} \circ \} \).[1]

c.Find

(i)     \(p \circ p\);

(ii)     the inverse of \(p \circ p\).[4]

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

(ii)     Give an example of an element with this order.[3]

▶️Answer/Explanation

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]

Question

Consider the set \(A\) consisting of all the permutations of the integers \(1,2,3,4,5\).

Two members of \(A\) are given by \(p = (1{\text{ }}2{\text{ }}5)\) and \(q = (1{\text{ }}3)(2{\text{ }}5)\).

a.Find the single permutation which is equivalent to \(q \circ p\).[4]

b.State a permutation belonging to \(A\) of order

(i)     \(4\);

(ii)     \(6\).[3]

c.Let \(P = \) {all permutations in \(A\) where exactly two integers change position},

and \(Q = \) {all permutations in \(A\) where the integer \(1\) changes position}.

(i)     List all the elements in \(P \cap Q\).

(ii)     Find \(n(P \cap Q’)\).[4]

▶️Answer/Explanation

Markscheme

\(q \circ p = (1{\text{ }}3)(2{\text{ }}5)(1{\text{ }}2{\text{ }}5)\)     (M1)

\( = (1{\text{ }}5{\text{ }}3)\)     M1A1A1

Note:     M1 for an answer consisting of disjoint cycles, A1 for \((1{\text{ }}5{\text{ }}3)\),

A1 for either \((2)\) or \((2)\) omitted.

Note:     Allow \(\left( {\begin{array}{*{20}{c}} 1&2&3&4&5 \\ 5&2&1&4&3 \end{array}} \right)\)

If done in the wrong order and obtained \((1{\text{ }}3{\text{ }}2)\), award A2.

[4 marks]

a.

(i)     any cycle with length \(4\) eg (\(1234\))     A1

(ii)     any permutation with \(2\) disjoint cycles one of length \(2\) and one of length \(3\) eg (\(1\) \(2\)) (\(3\) \(4\) \(5\))     M1A1

Note:     Award M1A0 for any permutation with \(2\) non-disjoint cycles one of length \(2\) and one of length \(3\).

Accept non cycle notation.

[3 marks]

b.

(i)     (\(1\), \(2\)), (\(1\), \(3\)), (\(1\), \(4\)), (\(1\), \(5\))     M1A1

(ii)     (\(2\) \(3\)), (\(2\) \(4\)), (\(2\) \(5\)), (\(3\) \(4\)), (\(3\) \(5\)), (\(4\) \(5\))     (M1)

6     A1

Note:     Award M1 for at least one correct cycle.

[4 marks]

Total [11 marks]

marks on this question.

b.

Many students were unable to start the question, seemingly as they did not understand the cyclic notation. Many of those that did understand found it quite straightforward to obtain good marks on this question.

c.
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