| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Groups with generators and relations |
| Difficulty | Challenging +1.3 This is a structured group theory question from FP3 requiring systematic application of given properties and table reading. Part (i) is a straightforward algebraic manipulation using commutativity. Parts (ii)-(iv) involve identifying subgroups from a provided Cayley table and checking isomorphism types—methodical but requiring careful verification rather than deep insight. While group theory is an advanced topic, the question guides students through each step with the table provided, making it moderately above average difficulty but not requiring novel problem-solving approaches. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03f Subgroups: definition and tests for proper subgroups8.03l Isomorphism: determine using informal methods |
| \(e\) | \(a\) | \(b\) | \(c\) | \(bc\) | \(ca\) | \(ab\) | \(abc\) | |
| \(e\) | \(e\) | \(a\) | \(b\) | \(c\) | \(bc\) | \(ca\) | \(ab\) | \(abc\) |
| \(a\) | \(a\) | \(e\) | \(ab\) | \(ca\) | \(abc\) | \(c\) | \(b\) | \(bc\) |
| \(b\) | \(b\) | \(ab\) | \(e\) | \(bc\) | \(c\) | \(abc\) | \(a\) | \(ca\) |
| \(c\) | \(c\) | \(ca\) | \(bc\) | \(e\) | \(b\) | \(a\) | \(abc\) | \(ab\) |
| \(bc\) | \(bc\) | \(abc\) | \(c\) | \(b\) | \(e\) | \(ab\) | \(ca\) | \(a\) |
| \(ca\) | \(ca\) | \(c\) | \(abc\) | \(a\) | \(ab\) | \(e\) | \(bc\) | \(b\) |
| \(ab\) | \(ab\) | \(b\) | \(a\) | \(abc\) | \(ca\) | \(bc\) | \(e\) | \(c\) |
| \(abc\) | \(abc\) | \(bc\) | \(ca\) | \(ab\) | \(a\) | \(b\) | \(c\) | \(e\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(abc = (ab)c = (ba)c = b(ac) = b(ca) = (bc)a = (cb)a = cba\) | M1 A1 2 | For using commutativity correctly; For correct proof (use of associativity may be implied) |
| Minimum working: \(abc = bac = bca = cba\) OR \(abc = acb = cab = cba\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{e, a\}, \{e, b\}, \{e, c\}, \{e, bc\}, \{e, ac\}, \{e, ab\}, \{e, abc\}\) | B1 B1 2 | For any 3 subgroups; For the other 2 subgroups and none incorrect... |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{e, a, b, ab\}, \{e, a, c, ca\}, \{e, b, c, bc\}\) | B1 | For any 3 subgroups |
| \(\{e, a, bc, abc\}, \{e, b, ca, abc\}, \{e, c, ab, abc\}\) | B1 | For 1 more subgroup |
| \(\{e, bc, ca, ab\}\) | B1 3 | For 1 more subgroup (5 in total) and none incorrect... |
| Answer | Marks | Guidance |
|---|---|---|
| All elements (\(\neq e\)) have order 2 | B1* | For appropriate reference to order of elements in \(G\) |
| OR all are self-inverse | ||
| OR no element of \(G\) has order 4 | ||
| OR no order 4 subgroup has a generator or is cyclic | ||
| OR subgroups are of the form \(\{e, a, b, ab\}\) (the Klein group) | ||
| \(\Rightarrow\) all order 4 subgroups are isomorphic | B1 (*dep)2 | For correct conclusion |
## (i)
$abc = (ab)c = (ba)c = b(ac) = b(ca) = (bc)a = (cb)a = cba$ | M1 A1 2 | For using commutativity correctly; For correct proof (use of associativity may be implied)
Minimum working: $abc = bac = bca = cba$ OR $abc = acb = cab = cba$ | |
## (ii)
$\{e, a\}, \{e, b\}, \{e, c\}, \{e, bc\}, \{e, ac\}, \{e, ab\}, \{e, abc\}$ | B1 B1 2 | For any 3 subgroups; For the other 2 subgroups and none incorrect...
## (iii)
$\{e, a, b, ab\}, \{e, a, c, ca\}, \{e, b, c, bc\}$ | B1 | For any 3 subgroups
$\{e, a, bc, abc\}, \{e, b, ca, abc\}, \{e, c, ab, abc\}$ | B1 | For 1 more subgroup
$\{e, bc, ca, ab\}$ | B1 3 | For 1 more subgroup (5 in total) and none incorrect...
## (iv)
All elements ($\neq e$) have order 2 | B1* | For appropriate reference to order of elements in $G$
OR all are self-inverse | |
OR no element of $G$ has order 4 | |
OR no order 4 subgroup has a generator or is cyclic | |
OR subgroups are of the form $\{e, a, b, ab\}$ (the Klein group) | |
$\Rightarrow$ all order 4 subgroups are isomorphic | B1 (*dep)2 | For correct conclusion
---A group $G$, of order 8, is generated by the elements $a$, $b$, $c$. $G$ has the properties
$$a^2 = b^2 = c^2 = e, \quad ab = ba, \quad bc = cb, \quad ca = ac,$$
where $e$ is the identity.
\begin{enumerate}[label=(\roman*)]
\item Using these properties and basic group properties as necessary, prove that $abc = cba$. [2]
\end{enumerate}
The operation table for $G$ is shown below.
\begin{center}
\begin{tabular}{c|cccccccc}
& $e$ & $a$ & $b$ & $c$ & $bc$ & $ca$ & $ab$ & $abc$ \\
\hline
$e$ & $e$ & $a$ & $b$ & $c$ & $bc$ & $ca$ & $ab$ & $abc$ \\
$a$ & $a$ & $e$ & $ab$ & $ca$ & $abc$ & $c$ & $b$ & $bc$ \\
$b$ & $b$ & $ab$ & $e$ & $bc$ & $c$ & $abc$ & $a$ & $ca$ \\
$c$ & $c$ & $ca$ & $bc$ & $e$ & $b$ & $a$ & $abc$ & $ab$ \\
$bc$ & $bc$ & $abc$ & $c$ & $b$ & $e$ & $ab$ & $ca$ & $a$ \\
$ca$ & $ca$ & $c$ & $abc$ & $a$ & $ab$ & $e$ & $bc$ & $b$ \\
$ab$ & $ab$ & $b$ & $a$ & $abc$ & $ca$ & $bc$ & $e$ & $c$ \\
$abc$ & $abc$ & $bc$ & $ca$ & $ab$ & $a$ & $b$ & $c$ & $e$ \\
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item List all the subgroups of order 2. [2]
\item List five subgroups of order 4. [3]
\item Determine whether all the subgroups of $G$ which are of order 4 are isomorphic. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q4 [9]}}