| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Complete or analyse Cayley table |
| Difficulty | Standard +0.8 This is a Further Maths group theory question requiring completion of a Cayley table using the Latin square property and identifying why the structure fails to be a group. While the Latin square completion is systematic, it requires careful logical deduction across multiple entries, and recognizing the specific group axiom failure (likely lack of identity or associativity) demands solid understanding of abstract algebra concepts that are beyond standard A-level. |
| Spec | 8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group |
| \(a\) | \(b\) | \(c\) | \(d\) | |
| \(a\) | \(b\) | \(a\) | ||
| \(b\) | ||||
| \(c\) | \(c\) | |||
| \(d\) | \(d\) | \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | a b c d |
| Answer | Marks |
|---|---|
| d d c b a All correct | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.1a |
| Answer | Marks |
|---|---|
| (b) | No – since c is the “left identity” but there is no “right identity” |
| Answer | Marks | Guidance |
|---|---|---|
| Or (ad)b = cb = b but a(db) = ac = a so op. not associative | B1 | |
| [1] | 2.4 | Any valid reason (vague or unsupported claims |
Question 1:
1 | (a) | a b c d
a b d a c Column 1
b c a d b Column 3
c a b c d Any 3rd R or C
d d c b a All correct | B1
B1
B1
B1
[4] | 3.1a
1.1
2.2a
1.1
(b) | No – since c is the “left identity” but there is no “right identity”
Or there is no identity column
Or (ad)b = cb = b but a(db) = ac = a so op. not associative | B1
[1] | 2.4 | Any valid reason (vague or unsupported claims
are not acceptable)1 The following Cayley table is for a set $\{ a , b , c , d \}$ under a suitable binary operation.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
& $a$ & $b$ & $c$ & $d$ \\
\hline
$a$ & $b$ & & $a$ & \\
\hline
$b$ & & & & \\
\hline
$c$ & & & $c$ & \\
\hline
$d$ & $d$ & & & $a$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Given that the Latin square property holds for this Cayley table, complete it using the table supplied in the Printed Answer Booklet.
\item Using your completed Cayley table, explain why the set does not form a group under the binary operation.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2020 Q1 [5]}}