| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Topic | Groups |
| Type | Prove group-theoretic identities |
| Difficulty | Challenging +1.8 This is a structured group theory proof requiring understanding of cancellation laws, permutation of group elements, and Lagrange's theorem. While the question guides students through the steps, it demands abstract reasoning about finite groups, careful justification of each claim, and understanding of why commutativity is essential. The multi-part structure and need to explain counterexamples places it well above average difficulty, though the scaffolding prevents it from being extremely challenging. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03k Lagrange's theorem: order of subgroup divides order of group |
**(i)(a)** $x \in G \Rightarrow \exists\, x^{-1} \in G$ and pre-multiplying by this (or $x$ in the $\Leftarrow$ case) gives the result. (NB **Both** directions must be dealt with) **B1 B1**
**[2]**
**(b)** Since each $xg_i$ is distinct, and there are $n$ of them, the set $xG$ is just a permutation of the elements of $G$ **OR** mention that it is just a row of the group table and hence contains a permutation of the elements of $G$ **B1**
**[1]**
**(ii)** Multiply all elements together: $xg_1\, xg_2\, xg_3\ldots xg_n = g_1\, g_2\, g_3\ldots g_n$ **E1**
(Since $G$ is abelian) $\Rightarrow x^n.(g_1\, g_2\, g_3\ldots g_n) = (g_1\, g_2\, g_3\ldots g_n)$ **E1**
Since $g_1\, g_2\, g_3\ldots g_n$ is an element of $G$, it has an inverse; Pre/post-mult. by this inverse then gives $x^n = e$ **E1**
**[3]**
**(iii)(a)** Elements may have an order which divides into (is a factor of) $n$ **B1**
**[1]**
**(b)** Because the change of the order of multiplications in $g.g_1\, g.g_2\, g.g_3\ldots g.g_n = g^n.(g_1\, g_2\, g_3\ldots g_n)$ is only valid in an abelian group **B1**
**[1]**8 Let $G = \left\{ g _ { 1 } , g _ { 2 } , g _ { 3 } , \ldots , g _ { n } \right\}$ be a finite abelian group of order $n$ under a multiplicative binary operation, where $g _ { 1 } = e$ is the identity of $G$.\\
(i) Let $x \in G$. Justify the following statements:
\begin{enumerate}[label=(\alph*)]
\item $x g _ { i } = x g _ { j } \Leftrightarrow g _ { i } = g _ { j }$;
\item $\left\{ x g _ { 1 } , x g _ { 2 } , x g _ { 3 } , \ldots , x g _ { n } \right\} = G$.\\
(ii) By considering the product of all $G$ 's elements, and using the result of part (i)(b), prove that $x ^ { n } = e$ for each $x \in G$.\\
(iii) Explain why\\
(a) this does not imply that all elements of $G$ have order $n$,\\
(b) this argument cannot be used to justify the same result for non-abelian groups.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2013 Q8 [8]}}