| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Verify group axioms |
| Difficulty | Standard +0.8 This is a substantial Further Maths group theory question requiring multiple techniques: algebraic manipulation to verify group axioms, proving closure using the factored form, finding the identity and inverses in S, proving no element has order 2, working with modular arithmetic to define a finite group, and identifying all subgroups. While each individual part is methodical rather than requiring deep insight, the length (7 parts), abstract nature of group theory, and need to systematically verify multiple properties places this above average difficulty for A-level. |
| Spec | 8.03a Binary operations: and their properties on given sets8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups |
| \(\circ\) | 0 | 1 | 2 | 4 | 5 | 6 |
| 0 | 0 | 1 | 2 | 4 | 5 | 6 |
| 1 | 1 | 4 | 0 | 6 | 2 | 5 |
| 2 | 2 | 0 | 5 | 1 | 6 | 4 |
| 4 | 4 | 6 | 1 | 5 | 0 | 2 |
| 5 | 5 | 2 | 6 | 0 | 4 | 1 |
| 6 | 6 | 5 | 4 | 2 | 1 | 0 |
## 4(i)
Commutative: $x * y = y * x$ (for all $x, y$)
**B1**
Associative: $(x * y) * z = x * (y * z)$ (for all $x, y, z$)
**B2**
**3** - Give B1 for a partial explanation, e.g. 'Position of brackets does not matter'
**Mark: 3**
## 4(ii)
$$2(x + \frac{1}{2})(y + \frac{1}{2}) - 1 = 2xy + x + y + \frac{1}{4} - 1 = 2xy + x + y - \frac{1}{4}$$
**B1 ag** - Intermediate step required
**Mark: 1**
## 4(iii)(A)
If $x, y \in S$ then $x > -\frac{1}{2}$ and $y > -\frac{1}{2}$
**M1**
$x + \frac{1}{2} > 0$ and $y + \frac{1}{2} > 0$, so $2(x + \frac{1}{2})(y + \frac{1}{2}) - 1 > -\frac{1}{2}$, so $x * y \in S$
**A1**
**A1**
**Mark: 3**
## 4(B)
0 is the identity since $0 * x = 0 + x = x$
**B1**
**B1**
If $x \in S$ and $x * y = 0$ then $2xy + x + y = 0$
**M1**
$$y = \frac{-x}{2x + 1}$$
**A1** - or $2(x + \frac{1}{2})(y + \frac{1}{2}) = \frac{1}{2}$ or $y + \frac{1}{2} = \frac{1}{4(x + \frac{1}{2})}$
$$y + \frac{1}{2} = \frac{1}{2(2(x + 1))} > 0 \quad (\text{since } x > -\frac{1}{2})$$
**M1**
so $y \in S$
**A1**
**Dependent on M1A1M1**
S is closed and associative; there is an identity; and every element of $S$ has an inverse in $S$
**Mark: 6**
## 4(iv)
If $x *4 A binary operation * is defined on real numbers $x$ and $y$ by
$$x * y = 2 x y + x + y$$
You may assume that the operation $*$ is commutative and associative.
\begin{enumerate}[label=(\roman*)]
\item Explain briefly the meanings of the terms 'commutative' and 'associative'.
\item Show that $x * y = 2 \left( x + \frac { 1 } { 2 } \right) \left( y + \frac { 1 } { 2 } \right) - \frac { 1 } { 2 }$.
The set $S$ consists of all real numbers greater than $- \frac { 1 } { 2 }$.
\item (A) Use the result in part (ii) to show that $S$ is closed under the operation *.\\
(B) Show that $S$, with the operation $*$, is a group.
\item Show that $S$ contains no element of order 2 .
The group $G = \{ 0,1,2,4,5,6 \}$ has binary operation ∘ defined by
$$x \circ y \text { is the remainder when } x * y \text { is divided by } 7 \text {. }$$
\item Show that $4 \circ 6 = 2$.
The composition table for $G$ is as follows.
\begin{center}
\begin{tabular}{ l | l l l l l l }
$\circ$ & 0 & 1 & 2 & 4 & 5 & 6 \\
\hline
0 & 0 & 1 & 2 & 4 & 5 & 6 \\
1 & 1 & 4 & 0 & 6 & 2 & 5 \\
2 & 2 & 0 & 5 & 1 & 6 & 4 \\
4 & 4 & 6 & 1 & 5 & 0 & 2 \\
5 & 5 & 2 & 6 & 0 & 4 & 1 \\
6 & 6 & 5 & 4 & 2 & 1 & 0 \\
\end{tabular}
\end{center}
\item Find the order of each element of $G$.
\item List all the subgroups of $G$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP3 2008 Q4 [24]}}