Show that ${(ba)^2} = e$ .

Express ${(bab)^{ - 1}}$ in its simplest form.

Given that $a \ne e$ ,

  (i)     show that $b \ne e$ ;

  (ii)     show that $G$ is not Abelian.

EITHER

$baba = ba{a^2}b$     M1

$= b{a^3}b$     (A1)

$= {b^2}$     A1

$= e$     AG

OR

$baba = {a^2}bba$     M1

$= {a^2}{b^2}a$      (A1)

$= {a^3}$     A1

$= e$     AG

[3 marks]

$bab = {a^2}bb$     (M1)

$= {a^2}$     (A1)

${(bab)^{ - 1}} = a$     A1

[3 marks]

(i)     assume $b = e$     M1

then $a = {a^2}$     A1

$\Rightarrow a = e$ which is a contradiction     R1

(ii)     if $ab = ba$     M1

then $ab = {a^2}b$     A1

$\Rightarrow a = e$ which is a contradiction     R1

[6 marks]

This question was started by the majority of candidates, but only successfully completed by a few. Many candidates seemed to be aware of this style of question, but were either unable to make significant progress or manipulated the algebra in a contorted manner and hence lost valuable time. Also a number of candidates made assumptions about commutativity which were not justified. Overall, the level and succinctness of meaningful algebraic manipulation shown by candidates was disappointing.

This question was started by the majority of candidates, but only successfully completed by a few. Many candidates seemed to be aware of this style of question, but were either unable to make significant progress or manipulated the algebra in a contorted manner and hence lost valuable time. Also a number of candidates made assumptions about commutativity which were not justified. Overall, the level and succinctness of meaningful algebraic manipulation shown by candidates was disappointing.

This question was started by the majority of candidates, but only successfully completed by a few. Many candidates seemed to be aware of this style of question, but were either unable to make significant progress or manipulated the algebra in a contorted manner and hence lost valuable time. Also a number of candidates made assumptions about commutativity which were not justified. In part (c) the idea of a proof by contradiction was used by stronger candidates, but weaker candidates were often at a loss as how to start. Overall, the level and succinctness of meaningful algebraic manipulation shown by candidates was disappointing.