| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2016 |
| Session | Specimen |
| Marks | 7 |
| Topic | Groups |
| Type | Prove group-theoretic identities |
| Difficulty | Challenging +1.8 This is a group theory question requiring knowledge of Lagrange's theorem, element orders, and algebraic manipulation of group equations. Part (i) is routine application of Lagrange's theorem. Parts (ii) and (iii) require non-trivial manipulation of the given relations and understanding of cyclic group structure, which is challenging but follows standard proof techniques for this level. The multi-step reasoning and abstract algebra content place it well above average A-level difficulty, though it's a structured question with clear conditions to work from rather than requiring deep novel insight. |
| Spec | 8.03e Order of elements: and order of groups8.03g Cyclic groups: meaning of the term8.03k Lagrange's theorem: order of subgroup divides order of group |
**(i)**
Possible orders are 1, 2, 3, 4, 6 & 12 **B1**
By *Lagrange's Theorem*, the order of an element divides the order of the group (since the order of an element $\equiv$ the order of the subgroup generated by that element) **B1**
**(ii)**
E.g. $y = yxy \Rightarrow y.x^2y = yxy.x^2y$ by ③ **M1**
$= xy.x^3.y = xy.y^2.y$ by ② **M1**
$= x.y^4 = x.(y^2)^2$ [by ②]
$= x.(x^3)^2 = x.e$ by ①
2 M's for first, correct uses of 2 different conditions; the **A** for the 3rd condition used to clinch the result. **A1**
**(iii)**
Proving $G$ not abelian: [e.g. by $xyx = y$ but $x^2 \neq e$] $\Rightarrow$ $G$ not cyclic **B1 B1**
**OR** establishing a contradiction
**Total: 7 marks**6 A group $G$ has order 12.\\
(i) State, with a reason, the possible orders of the elements of $G$.
The identity element of $G$ is $e$, and $x$ and $y$ are distinct, non-identity elements of $G$ satisfying the three conditions\\
(1) $x$ has order 6 ,\\
(2) $x ^ { 3 } = y ^ { 2 }$,\\
(3) $x y x = y$.\\
(ii) Prove that $y x ^ { 2 } y = x$.\\
(iii) Prove that $G$ is not a cyclic group.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q6 [7]}}