| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | December |
| Marks | 12 |
| Topic | Groups |
| Type | Verify group axioms |
| Difficulty | Challenging +1.2 This is a systematic group axioms verification question requiring multiple steps (closure, associativity, identity, inverse) plus finding a subgroup. While it involves abstract algebra (Further Maths territory), each part follows standard procedures: closure and associativity are algebraic verification, identity/inverse require solving equations. The subgroup part is straightforward once the inverse is known. More demanding than typical A-level due to the abstract nature and length, but doesn't require deep insight—just careful algebraic manipulation following learned procedures. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03g Cyclic groups: meaning of the term |
| Answer | Marks | Guidance |
|---|---|---|
| Since \(a \neq 0\) and \(c \neq 0, ac \neq 0 \Rightarrow (ac, b + ad) \in L\) so \(L\) is closed under \(*\) | B1 | 2.4 |
| Total: [1] |
| Answer | Marks | Guidance |
|---|---|---|
| \([(a,b) \ominus (c,d)] \ominus (e,f) = (ac, b + ad) \ominus (e,f)\) | M1 | 1.1a |
| \(= (ace, b + ad + acf)\) | A1 | 1.1 |
| \((a,b) \ominus [(c,d) \ominus (e,f)] = (a,b) \ominus (ce, d + cf)\) | M1 | 2.1 |
| \(= (ace, b + ad(d + cf))\) | A1 | 2.4 |
| Since \(a(d + cf) = ad + acf\) the two are equal, we have associativity | A1 | 2.4 |
| Total: [4] | Attempt at \((x\ominus y)\ominus z\) or \(x\ominus(y\ominus z)\) for elements \(x, y, z\); The other attempted; Must state that the two are equal |
| Answer | Marks | Guidance |
|---|---|---|
| \((a,b) \ominus (c,d) = (ac, b + ad) = (a,b)\) e.g. | M1 | 1.1a |
| \(\Rightarrow c = 1, d = 0\) i.e. \((1,0)\) is the identity | A1 | 1.1 |
| Total: [2] | Setting product equal to either component |
| Answer | Marks | Guidance |
|---|---|---|
| \((a,b) \ominus (c,d) = (ac, b + ad) = (1,0)\) | M1 | 1.1a |
| gives \(c = \frac{1}{a}\) and \(d = -\frac{b}{a}\) so that \((a,b)^{-1} = \left(\frac{1}{a}, -\frac{b}{a}\right)\) | A1 | 1.1 |
| which exists and is in \(L\) since \(a \neq 0\) and \(\frac{1}{a} \neq 0\) | E1 | 2.4 |
| Total: [3] | Must note here that the inverse is in \(L\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((a,b) * (a,b) = (1,0)\) i.e. \((a^2, b(1+a)) = (1,0)\) | M1 | 3.1a |
| For this not to be the identity, \(a = -1\) and \(b\) arbitrary | A1 | 2.2a |
| So the set \(\{(1,0), (-1,b)\}\) for any real \(b\) will form a subgroup of order 2 | A1 | 2.2a |
| Total: [2] | Looking for any self-inverse element; Must choose a value for \(b\) or note that any will do |
**Part (a)(i)**
| Since $a \neq 0$ and $c \neq 0, ac \neq 0 \Rightarrow (ac, b + ad) \in L$ so $L$ is closed under $*$ | B1 | 2.4 |
| Total: [1] | | |
**Part (a)(ii)**
| $[(a,b) \ominus (c,d)] \ominus (e,f) = (ac, b + ad) \ominus (e,f)$ | M1 | 1.1a |
| $= (ace, b + ad + acf)$ | A1 | 1.1 |
| $(a,b) \ominus [(c,d) \ominus (e,f)] = (a,b) \ominus (ce, d + cf)$ | M1 | 2.1 |
| $= (ace, b + ad(d + cf))$ | A1 | 2.4 |
| Since $a(d + cf) = ad + acf$ the two are equal, we have associativity | A1 | 2.4 |
| Total: [4] | Attempt at $(x\ominus y)\ominus z$ or $x\ominus(y\ominus z)$ for elements $x, y, z$; The other attempted; Must state that the two are equal | |
**Part (a)(iii)**
| $(a,b) \ominus (c,d) = (ac, b + ad) = (a,b)$ e.g. | M1 | 1.1a |
| $\Rightarrow c = 1, d = 0$ i.e. $(1,0)$ is the identity | A1 | 1.1 |
| Total: [2] | Setting product equal to either component | |
**Part (a)(iv)**
| $(a,b) \ominus (c,d) = (ac, b + ad) = (1,0)$ | M1 | 1.1a |
| gives $c = \frac{1}{a}$ and $d = -\frac{b}{a}$ so that $(a,b)^{-1} = \left(\frac{1}{a}, -\frac{b}{a}\right)$ | A1 | 1.1 |
| which exists and is in $L$ since $a \neq 0$ and $\frac{1}{a} \neq 0$ | E1 | 2.4 |
| Total: [3] | Must note here that the inverse is in $L$ | |
**Part (b)**
| $(a,b) * (a,b) = (1,0)$ i.e. $(a^2, b(1+a)) = (1,0)$ | M1 | 3.1a |
| For this not to be the identity, $a = -1$ and $b$ arbitrary | A1 | 2.2a |
| So the set $\{(1,0), (-1,b)\}$ for any real $b$ will form a subgroup of order 2 | A1 | 2.2a |
| Total: [2] | Looking for any self-inverse element; Must choose a value for $b$ or note that any will do | |4 The set $L$ consists of all points $( x , y )$ in the cartesian plane, with $x \neq 0$. The operation ◇ is defined by $( a , b ) \diamond ( c , d ) = ( a c , b + a d )$ for $( a , b ) , ( c , d ) \in L$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $L$ is closed under ◇.
\item Prove that $\diamond$ is associative on $L$.
\item Find the identity element of $L$ under ◇ .
\item Find the inverse element of $( a , b )$ under ◇.
\end{enumerate}\item Find a subgroup of $( L , \diamond )$ of order 2.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q4 [12]}}