Pre-U Pre-U 9795/1 2012 June — Question 6 7 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2012
SessionJune
Marks7
TopicGroups
TypeProve group-theoretic identities
DifficultyChallenging +1.8 This is a group theory question requiring knowledge of Lagrange's theorem and multi-step algebraic manipulation with group elements. Part (i) is routine application of Lagrange's theorem. Parts (ii) and (iii) require careful symbolic manipulation using the given relations—non-trivial but systematic work rather than deep insight. This is challenging for A-level but standard for Further Mathematics group theory.
Spec8.03e 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

6 A group \(G\) has order 12.
  1. 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) \(\quad x\) has order 6 ,
    (2) \(x ^ { 3 } = y ^ { 2 }\),
    (3) \(x y x = y\).
  2. Prove that \(y x ^ { 2 } y = x\).
  3. Prove that \(G\) is not a cyclic 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 ≡ the order of the subgroup generated by that element) B1
[2]
(ii) E.g. \(y = xyx \Rightarrow y\cdot x^2y = xyx\cdot x^2y\) by ③ M1
\(= xy\cdot x^3\cdot y = xy\cdot y^2\cdot y\) by ② M1
\(= x\cdot y^4 = x\cdot(y^2)^2\)
\(= x\cdot(x^3)^2 = x\cdot e\) 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
SPECIAL CASE Allow 2/3 for those who correctly argue the converse
[3]
(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
[2]
**(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 ≡ the order of the subgroup generated by that element) **B1**

**[2]**

**(ii)** E.g. $y = xyx \Rightarrow y\cdot x^2y = xyx\cdot x^2y$ by ③ **M1**

$= xy\cdot x^3\cdot y = xy\cdot y^2\cdot y$ by ② **M1**

$= x\cdot y^4 = x\cdot(y^2)^2$

$= x\cdot(x^3)^2 = x\cdot e$ 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**

**SPECIAL CASE** Allow 2/3 for those who correctly argue the converse

**[3]**

**(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

**[2]**
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) $\quad x$ 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 2012 Q6 [7]}}