| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Isomorphism between groups |
| Difficulty | Challenging +1.2 This is a standard Further Maths group theory question requiring construction of an operation table, verification of group axioms, finding element orders and subgroups, and checking isomorphism. While it involves multiple parts and requires systematic work, these are routine techniques for FP3 students with no novel insights needed. The modulo 10 multiplication makes calculations straightforward, and isomorphism checking follows standard procedures (comparing orders/structure). |
| Spec | 8.03a Binary operations: and their properties on given sets8.03b Cayley tables: construct for finite sets under binary operation8.03c Group definition: recall and use, show structure is/isn't a group8.03e Order of elements: and order of groups8.03f Subgroups: definition and tests for proper subgroups8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Cayley table with rows/columns \(1,3,5,7\) (mod 8) correctly completed | B2 | \(-1\) each error |
| From table clearly closed | B1 | Must be clear they are referring to tabulated results |
| 1 is identity | B1 | Or "1 appears in every row" |
| \(3^{-1} \equiv 3,\ 5^{-1} \equiv 5,\ 7^{-1} \equiv 7 \pmod{8}\) | B1 | Superfluous facts gets \(-1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 1 has order 1 and 3, 5, 7 all have order 2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\{1,3\},\ \{1,5\},\ \{1,7\}\) (and \(\{1\}\)) | B1 | All correct, no extras. Allow \(\{1\}\) included or omitted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| In \(H\): \(3^2 \equiv 9 \pmod{10}\) so 3 not order 2 | M1 | Shows and states that 3 or that 7 is not order 2 (or is order 4) |
| No element of order \(> 2\) in \(G\) so not isomorphic | A1 | Completely correct reasoning. Or, if zero, SC1 for merely stating comparable orders and then saying "orders don't correspond, so not isomorphic". Or table for H with saying "not all elements self inverse, so not isomorphic" |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Cayley table with rows/columns $1,3,5,7$ (mod 8) correctly completed | B2 | $-1$ each error |
| From table clearly closed | B1 | Must be clear they are referring to tabulated results |
| 1 is identity | B1 | Or "1 appears in every row" |
| $3^{-1} \equiv 3,\ 5^{-1} \equiv 5,\ 7^{-1} \equiv 7 \pmod{8}$ | B1 | Superfluous facts gets $-1$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 1 has order 1 and 3, 5, 7 all have order 2 | B1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\{1,3\},\ \{1,5\},\ \{1,7\}$ (and $\{1\}$) | B1 | All correct, no extras. Allow $\{1\}$ included or omitted |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| In $H$: $3^2 \equiv 9 \pmod{10}$ so 3 not order 2 | M1 | Shows and states that 3 or that 7 is not order 2 (or is order 4) |
| No element of order $> 2$ in $G$ so not isomorphic | A1 | Completely correct reasoning. Or, if zero, SC1 for merely stating comparable orders and then saying "orders don't correspond, so not isomorphic". Or table for H with saying "not all elements self inverse, so not isomorphic" |
---(i) Write down the operation table and, assuming associativity, show that $G$ is a group.\\
(ii) State the order of each element.\\
(iii) Find all the proper subgroups of $G$.
The group $H$ consists of the set $\{ 1,3,7,9 \}$ with the operation of multiplication modulo 10 .\\
(iv) Explaining your reasoning, determine whether $H$ is isomorphic to $G$.
\hfill \mbox{\textit{OCR FP3 2013 Q2 [9]}}