Show that R is an equivalence relation.

The Cayley table for G is shown below.

The subgroup H is given as $H = \{ e,{\text{ }}{a^2}b\}$.

(i)     Find the equivalence class with respect to R which contains ab.

(ii)     Another equivalence relation $\rho$ is defined on G by $x\rho y$ if and only if ${x^{ - 1}}y \in H$, for $x,{\text{ }}y \in G$. Find the equivalence class with respect to $\rho$ which contains ab.

$x{x^{ - 1}} = e \in H$     M1

$\Rightarrow xRx$

hence R is reflexive     A1

if xRy then $x{y^{ - 1}} \in H$

$\Rightarrow {(x{y^{ - 1}})^{ - 1}} \in H$     M1

now $(x{y^{ - 1}}){(x{y^{ - 1}})^{ - 1}} = e$ and $x{y^{ - 1}}y{x^{ - 1}} = e$

$\Rightarrow {(x{y^{ - 1}})^{ - 1}} = y{x^{ - 1}}$     A1

hence $y{x^{ - 1}} \in H \Rightarrow yRx$

hence R is symmetric     A1

if xRy, yRz then $x{y^{ - 1}} \in H,{\text{ }}y{z^{ - 1}} \in H$     M1

$\Rightarrow (x{y^{ - 1}})(y{z^{ - 1}}) \in H$     M1

$\Rightarrow x({y^{ - 1}}y){z^{ - 1}} \in H$

$\Rightarrow {x^{ - 1}}z \in H$

hence R is transitive     A1

hence R is an equivalence relation     AG

[8 marks]

(i)     for the equivalence class, solving:

EITHER

$x{(ab)^{ - 1}} = e{\text{ or }}x{(ab)^{ - 1}} = {a^2}b$     (M1)

$\{ ab,{\text{ }}a\}$     A2

OR

$ab{(x)^{ - 1}} = e{\text{ or }}ab{(x)^{ - 1}} = {a^2}b$     (M1)

$\{ ab,{\text{ }}a\}$     A2

(ii)     for the equivalence class, solving:

EITHER

${x^{ - 1}}(ab) = e{\text{ or }}{x^{ - 1}}(ab) = {a^2}b$     (M1)

$\{ ab,{\text{ }}{a^2}\}$     A2

OR

${(ab)^{ - 1}}x = e{\text{ or }}{(ab)^{ - 1}}x = {a^2}b$     (M1)

$\{ ab,{\text{ }}{a^2}\}$     A2 

[6 marks]

Stronger candidates made a reasonable start to (a), and many were able to demonstrate that the relation was reflexive and transitive. However, the majority of candidates struggled to make a meaningful attempt to show the relation was symmetric, with many making unfounded assumptions. Equivalence classes still cause major problems and few fully correct answers were seen to (b).

Stronger candidates made a reasonable start to (a), and many were able to demonstrate that the relation was reflexive and transitive. However, the majority of candidates struggled to make a meaningful attempt to show the relation was symmetric, with many making unfounded assumptions. Equivalence classes still cause major problems and few fully correct answers were seen to (b).