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.

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]