| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Prove group-theoretic identities |
| Difficulty | Challenging +1.2 Part (i) is a standard group theory identity requiring direct application of inverse properties. Part (ii) requires algebraic manipulation using the given condition and group axioms, involving several steps but following a clear logical path. Part (iii) tests understanding by asking for a counterexample or proof, which requires some insight. While this is Further Maths content (inherently harder), these are relatively routine group theory proofs that follow standard techniques taught in FP3, making it moderately above average difficulty overall. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1 \((yx)(yx)^{-1} = e \Rightarrow x(yx)^{-1} = y^{-1}\) | M1 | For starting point and appropriate multiplication |
| \(\Rightarrow (yxx)^{-1} = x^{-1}y^{-1}\) | A1 | For correct result AG |
| [2] | ||
| METHOD 2 Compare \((yx)(yx)^{-1} = e\) with \(yxx^{-1}y^{-1} = e\) | M1 | For appropriate comparison |
| \(\Rightarrow (yx)^{-1} = x^{-1}y^{-1}\) | A1 | For correct result AG For A1, proof cannot be written in the form 'LHS = RHS \(\rightarrow\) ... \(\rightarrow\) e = \(e\)' |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^ny^n = (xy)^n = x(yx)^{n-1}y\) | M1 | For using associativity or an inverse with respect to LHS, RHS or initial equality www beforehand |
| \(\Rightarrow x^{-1}x^ny^n y^{-1} = x^{-1}x(yx)^{n-1}yy^{-1}\) | M1 | For using \((xy)^n = x(yx)^{n-1}y\) oe |
| \(\Rightarrow x^{n-1}y^{n-1} = (yx)^{n-1}\) | A1 | For correct result AG SR for numerical \(n\) used, allow M1 M1 only |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1 All steps in (ii) are reversible \(\Rightarrow\) result follows | B1*dep B1*dep | For correct reason. Dep on correct part(ii) For correct conclusion |
| [2] | ||
| METHOD 2 Show working for (ii) in reverse \(\Rightarrow\) result follows | B1* B1*dep | For correct working For correct conclusion |
**(i)**
| METHOD 1 $(yx)(yx)^{-1} = e \Rightarrow x(yx)^{-1} = y^{-1}$ | M1 | For starting point and appropriate multiplication |
| $\Rightarrow (yxx)^{-1} = x^{-1}y^{-1}$ | A1 | For correct result AG |
| | [2] | |
| METHOD 2 Compare $(yx)(yx)^{-1} = e$ with $yxx^{-1}y^{-1} = e$ | M1 | For appropriate comparison |
| $\Rightarrow (yx)^{-1} = x^{-1}y^{-1}$ | A1 | For correct result AG For A1, proof cannot be written in the form 'LHS = RHS $\rightarrow$ ... $\rightarrow$ e = $e$' |
**(ii)**
| $x^ny^n = (xy)^n = x(yx)^{n-1}y$ | M1 | For using associativity or an inverse with respect to LHS, RHS or initial equality www beforehand |
| $\Rightarrow x^{-1}x^ny^n y^{-1} = x^{-1}x(yx)^{n-1}yy^{-1}$ | M1 | For using $(xy)^n = x(yx)^{n-1}y$ oe |
| $\Rightarrow x^{n-1}y^{n-1} = (yx)^{n-1}$ | A1 | For correct result AG SR for numerical $n$ used, allow M1 M1 only |
| | [3] | |
**(iii)**
| METHOD 1 All steps in (ii) are reversible $\Rightarrow$ result follows | B1*dep B1*dep | For correct reason. Dep on correct part(ii) For correct conclusion |
| | [2] | |
| METHOD 2 Show working for (ii) in reverse $\Rightarrow$ result follows | B1* B1*dep | For correct working For correct conclusion |3 A multiplicative group contains the distinct elements $e , x$ and $y$, where $e$ is the identity.\\
(i) Prove that $x ^ { - 1 } y ^ { - 1 } = ( y x ) ^ { - 1 }$.\\
(ii) Given that $x ^ { n } y ^ { n } = ( x y ) ^ { n }$ for some integer $n \geqslant 2$, prove that $x ^ { n - 1 } y ^ { n - 1 } = ( y x ) ^ { n - 1 }$.\\
(iii) If $x ^ { n - 1 } y ^ { n - 1 } = ( y x ) ^ { n - 1 }$, does it follow that $x ^ { n } y ^ { n } = ( x y ) ^ { n }$ ? Give a reason for your answer.
\hfill \mbox{\textit{OCR FP3 2012 Q3 [7]}}