Date | May 2018 | Marks available | 2 | Reference code | 18M.3srg.hl.TZ0.4 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | Determine | Question number | 4 | Adapted from | N/A |
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.
Find the proper subgroup H of order 6 containing \({p_1}\), \({p_2}\) and their compositions. Express each element of H in cycle form.
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.
Markscheme
number of possible permutations is 4 × 3 × 2 × 1 (M1)
= 24(= 4!) A1
[2 marks]
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]
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]