| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Subgroups and cosets |
| Difficulty | Challenging +1.8 This question requires proving H is a subgroup using closure, identity, and inverses (standard but requires careful abstract reasoning), then systematically checking commutativity in a Cayley table. Part (i) is a formal proof about the center of a group, requiring understanding of subgroup axioms in an abstract setting. Part (ii) involves methodical verification but is computational. The abstract nature and proof requirement elevate this above typical A-level, though the techniques are accessible to strong FM students. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03f Subgroups: definition and tests for proper subgroups |
| \(e\) | \(p\) | \(q\) | \(r\) | \(s\) | \(t\) | |
| \(e\) | \(e\) | \(p\) | \(q\) | \(r\) | \(s\) | \(t\) |
| \(p\) | \(p\) | \(q\) | \(e\) | \(s\) | \(t\) | \(r\) |
| \(q\) | \(q\) | \(e\) | \(p\) | \(t\) | \(r\) | \(s\) |
| \(r\) | \(r\) | \(t\) | \(s\) | \(e\) | \(q\) | \(p\) |
| \(s\) | \(s\) | \(r\) | \(t\) | \(p\) | \(e\) | \(q\) |
| \(t\) | \(t\) | \(s\) | \(r\) | \(q\) | \(p\) | \(e\) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(eg = ge\) so \(e \in H\) | B1 | Showing identity in \(H\) |
| \(hg = gh \Rightarrow g = h^{-1}gh \Rightarrow gh^{-1} = h^{-1}g\) | M1 | |
| \(\Rightarrow gh^{-1} = h^{-1}g\) | M1 | |
| \(\Rightarrow h^{-1} \in H\) | A1 | |
| \(h_1h_2g = h_1 gh_2 = gh_1h_2\) so \(h_1h_2 \in H\), so \(H\) closed | M1 | |
| M1 | ||
| so \(H_1h_2 \in H\), so \(H\) closed | A1 | |
| so \(H\) is a subgroup of \(G\) | A1 | For completing argument without considering other properties of \(H\). |
| [8] | ||
| (ii) Correctly evaluates first \(g_1 g_2\) \(g_1 g_2 \neq g_2 g_1\) for one correct pair | B1* | where \(g_1, g_2\) distinct and \(\neq e\) |
| \(g_1 g_2 \neq g_2 g_1\) for sufficient pairs to cover all 5 elements and conclude that they are not in \(H\) so \(H = \{e\}\) | A1 | |
| A1dep* | Complete argument | |
| [4] |
**(i)** $eg = ge$ so $e \in H$ | B1 | Showing identity in $H$
$hg = gh \Rightarrow g = h^{-1}gh \Rightarrow gh^{-1} = h^{-1}g$ | M1 |
$\Rightarrow gh^{-1} = h^{-1}g$ | M1 |
$\Rightarrow h^{-1} \in H$ | A1 |
$h_1h_2g = h_1 gh_2 = gh_1h_2$ so $h_1h_2 \in H$, so $H$ closed | M1 |
| M1 |
so $H_1h_2 \in H$, so $H$ closed | A1 |
so $H$ is a subgroup of $G$ | A1 | For completing argument without considering other properties of $H$.
| [8] |
**(ii)** Correctly evaluates first $g_1 g_2$ $g_1 g_2 \neq g_2 g_1$ for one correct pair | B1* | where $g_1, g_2$ distinct and $\neq e$
$g_1 g_2 \neq g_2 g_1$ for sufficient pairs to cover all 5 elements and conclude that they are not in $H$ so $H = \{e\}$ | A1 |
| A1dep* | Complete argument
| [4] |8 Let $G$ be any multiplicative group. $H$ is a subset of $G$. $H$ consists of all elements $h$ such that $h g = g h$ for every element $g$ in $G$.\\
(i) Prove that $H$ is a subgroup of $G$.
Now consider the case where $G$ is given by the following table:
\begin{center}
\begin{tabular}{ c | c c c c c c }
& $e$ & $p$ & $q$ & $r$ & $s$ & $t$ \\
\hline
$e$ & $e$ & $p$ & $q$ & $r$ & $s$ & $t$ \\
$p$ & $p$ & $q$ & $e$ & $s$ & $t$ & $r$ \\
$q$ & $q$ & $e$ & $p$ & $t$ & $r$ & $s$ \\
$r$ & $r$ & $t$ & $s$ & $e$ & $q$ & $p$ \\
$s$ & $s$ & $r$ & $t$ & $p$ & $e$ & $q$ \\
$t$ & $t$ & $s$ & $r$ & $q$ & $p$ & $e$ \\
\end{tabular}
\end{center}
(ii) Show that $H$ consists of just the identity element.
\hfill \mbox{\textit{OCR FP3 2015 Q8 [12]}}