Challenging +1.8 This is a Further Maths group theory question requiring students to construct part of a group table and use closure to derive a non-obvious identity. It demands understanding of group axioms, orders of elements, and systematic case-by-case reasoning—significantly harder than standard A-level content but more guided than open-ended proofs.
3 A non-commutative group \(G\) consists of the six elements \(\left\{ e , a , a ^ { 2 } , b , a b , b a \right\}\) where \(e\) is the identity element, \(a\) is an element of order 3 and \(b\) is an element of order 2 .
By considering the row in \(G\) 's group table in which each of the above elements is pre-multiplied by \(b\), show that \(b a ^ { 2 } = a b\).
Question 3:
3 | bG = {be, ba, ba2, b2, bab, b2a}
= { b, ba, ??, e, ??, a } using b2 = e
So ba2 is either a2 or ab
But ba2 = a2 only if b = e, which it isn’t
so ba2 = ab | M1
A1
A1
M1
E1
[5] | 1.1a
1.1
1.1
2.1
2.2a | Attempt to pre-multiply by b
All six elements, unsimplified
Noting that four elements are
known
Use of the Latin-Square property
3 A non-commutative group $G$ consists of the six elements $\left\{ e , a , a ^ { 2 } , b , a b , b a \right\}$ where $e$ is the identity element, $a$ is an element of order 3 and $b$ is an element of order 2 .\\
By considering the row in $G$ 's group table in which each of the above elements is pre-multiplied by $b$, show that $b a ^ { 2 } = a b$.
\hfill \mbox{\textit{OCR Further Additional Pure AS Q3 [5]}}