| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 2 |
| Topic | Groups |
| Type | Prove group-theoretic identities |
| Difficulty | Hard +2.3 This is a sophisticated group theory question requiring multiple proof techniques (contradiction, counterexample construction), deep understanding of group axioms and order properties, and the ability to work with abstract structures. Part (iv) requires particularly clever reasoning about cosets and element orders. This is significantly harder than standard A-level content and requires the mathematical maturity expected in Further Mathematics extension modules. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups |
| Answer | Marks | Guidance |
|---|---|---|
| If \( | G | = 2k\), then there are \(2k-1\) elements to pair up |
(i) All non-identity elements pair up with their inverses.
If $|G| = 2k$, then there are $2k-1$ elements to pair up
$\Rightarrow$ at least one element "pairs up" with itself. **M1 A1** [2]
(ii) $ab = (ab)^{-1} = b^{-1}a^{-1} = ba$ **M1 A1** [2]
**ALT.** $(ab)^2 = i \Rightarrow abab = i \Rightarrow aba = b$ (post-multiplying by $b = b^{-1}$)
$\Rightarrow ab = ba$ (post-multiplying by $a = a^{-1}$)
(iii) E.g. Let $x = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}$ and $y = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix}$, each of order 2 **B1**
Then $xy = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \\ 2 & 3 & 1 \end{pmatrix}$ (transforming) OR $\begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \\ 3 & 1 & 2 \end{pmatrix}$ (transposing) of order 3 **B1** [2]
(iv) **(a)** $H$ closed since $x(yx) = x(xy) = x^2y = y$
and $y(xy) = y^2x = x$ using result of **(ii)** **M1 A1**
[Associativity follows from that of $G$]
The identity is in $H$ and each element has its own inverse (itself) in $H$ **B1**
Hence $H$ a subgroup of $G$ **B1**
However, Lagrange's Theorem states that $o(H) \mid o(G)$ and $4 \mid 4n+2$
contradicting the assumption that $G$ can contain all self-inverse elements. **E1** [5]9 (i) Explain why all groups of even order must contain at least one self-inverse element (that is, an element of order 2).\\
(ii) Prove that any group in which every non-identity element is self-inverse is abelian.\\
(iii) Simon believes that if $x$ and $y$ are two distinct self-inverse elements of a group, then the element $x y$ is also self-inverse. By considering the group of the six permutations of $\left( \begin{array} { l l } 1 & 2 \end{array} \right)$, produce a counter-example to prove him wrong.\\
(iv) 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 self-inverse elements of $G$. By considering the set $H = \{ i , x , y , x y \}$, prove by contradiction that $G$ cannot contain all self-inverse elements.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q9 [2]}}