| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Prove group-theoretic identities |
| Difficulty | Challenging +1.8 This is a structured group theory question requiring multiple proofs. Part (a) uses a pairing argument (moderate), part (b) requires algebraic manipulation to prove commutativity (standard technique), part (c) is straightforward verification from a given table, and part (d) requires a contradiction proof using order arguments and Lagrange's theorem. While conceptually accessible for Further Maths students who have studied group theory, it demands careful reasoning across multiple parts and familiarity with proof techniques, placing it well above average difficulty. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups |
| \(i\) | \(a\) | \(b\) | \(c\) | \(d\) | \(e\) | |
| \(i\) | \(i\) | \(a\) | \(b\) | \(c\) | \(d\) | \(e\) |
| \(a\) | \(a\) | \(i\) | \(d\) | \(e\) | \(b\) | \(c\) |
| \(b\) | \(b\) | \(e\) | \(i\) | \(d\) | \(c\) | \(a\) |
| \(c\) | \(c\) | \(d\) | \(e\) | \(i\) | \(a\) | \(b\) |
| \(d\) | \(d\) | \(c\) | \(a\) | \(b\) | \(e\) | \(i\) |
| \(e\) | \(e\) | \(b\) | \(c\) | \(a\) | \(i\) | \(d\) |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | a | Inverses pair up (even number) and there is one identity |
| Answer | Marks |
|---|---|
| inverse elements | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.1 |
| 2.2a | Parity argument involving inverse pairs |
| b | Consider xy and its inverse, also xy. |
| Answer | Marks |
|---|---|
| = y– 1x– 1 = yx ( abelian) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | Must be stated or derived |
| Answer | Marks |
|---|---|
| c | By considering the product of any two of elements a, b, c |
| Answer | Marks |
|---|---|
| … gives one of d, e of order 3 | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.1 |
| 2.2a | It must be noted that these elements are self-inverse (or |
| Answer | Marks |
|---|---|
| d | Assume that all (non-identity) elements of G are self- |
| Answer | Marks |
|---|---|
| (Lagrange’s theorem states that) o(H) | o(G) |
| But 4 ∤ (4n + 2) giving the required | B1 |
| Answer | Marks |
|---|---|
| [4] | 2.1 |
| Answer | Marks |
|---|---|
| 1.1 | Longer proofs that H is a group may involve xy = yx |
Question 8:
8 | a | Inverses pair up (even number) and there is one identity
element
Since total is even, there must be an odd number of self-
inverse elements | M1
A1
[2] | 2.1
2.2a | Parity argument involving inverse pairs
b | Consider xy and its inverse, also xy.
Now xy = (xy)– 1
= y– 1x– 1 = yx ( abelian) | M1
A1
[2] | 3.1a
1.1 | Must be stated or derived
NB No need to include statements about the identity
c | By considering the product of any two of elements a, b, c
…
… gives one of d, e of order 3 | M1
A1
[2] | 2.1
2.2a | It must be noted that these elements are self-inverse (or
of order 2)
Correct example from the Cayley table must be given
and the contradiction concluded
d | Assume that all (non-identity) elements of G are self-
inverse/have order 2
Then H is a group ( the Klein group of order 4) and hence
a subgroup of G
(Lagrange’s theorem states that) o(H) | o(G)
But 4 ∤ (4n + 2) giving the required | B1
B1*
M1
A1dep
[4] | 2.1
2.2a
2.4
1.1 | Longer proofs that H is a group may involve xy = yx
(by part b) so that H is closed, etc.
Consideration of Lagrange’s theorem8
\begin{enumerate}[label=(\alph*)]
\item Explain why all groups of even order must contain at least one self-inverse element (that is, an element of order 2).
\item Prove that any group, in which every (non-identity) element is self-inverse, is abelian.
\item A student believes that, if $x$ and $y$ are two distinct, non-identity, self-inverse elements of a group, then the element $x y$ is also self-inverse.
The table shown here is the Cayley table for the non-cyclic group of order 6, having elements $i , a , b , c , d$ and $e$, where $i$ is the identity.
\begin{center}
\begin{tabular}{ c | | c | c | c | c | c | c | }
& $i$ & $a$ & $b$ & $c$ & $d$ & $e$ \\
\hline\hline
$i$ & $i$ & $a$ & $b$ & $c$ & $d$ & $e$ \\
\hline
$a$ & $a$ & $i$ & $d$ & $e$ & $b$ & $c$ \\
\hline
$b$ & $b$ & $e$ & $i$ & $d$ & $c$ & $a$ \\
\hline
$c$ & $c$ & $d$ & $e$ & $i$ & $a$ & $b$ \\
\hline
$d$ & $d$ & $c$ & $a$ & $b$ & $e$ & $i$ \\
\hline
$e$ & $e$ & $b$ & $c$ & $a$ & $i$ & $d$ \\
\hline
\end{tabular}
\end{center}
By considering the elements of this group, produce a counter-example which proves that this student is wrong.
\item A group $G$ has order $4 n + 2$, for some positive integer $n$, and $i$ is the identity element of $G$. Let $x$ and $y$ be two distinct, non-identity, self-inverse elements of $G$. By considering the set $\mathrm { H } = \{ \mathrm { i } , \mathrm { x } , \mathrm { y } , \mathrm { xy } \}$, prove by contradiction that not all elements of $G$ are self-inverse.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2022 Q8 [10]}}