| Answer | Marks | Guidance |
|---|---|---|
| Translates the problem to that of finding a generator of \(C\) | M1 | AO 3.1a |
| Shows that 1 (or 3) is a generator of \(C\) and concludes \(C\) is cyclic | A1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| States the rotations of the square | B1 | AO 2.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Sets up the condition for an element and its inverse: \((-1, 1) \otimes (a, b) = (1, 1)\), so \((-a, b) = (1, 1)\) | M1 | AO 1.1a |
| Identifies the correct inverse of \((-1, 1)\): Therefore \(a = -1\) and \(b = 1\), so the inverse of \((-1, 1)\) is \((-1, 1)\) | A1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(V\) does not contain a generator as all elements in \(V\) are self-inverse. Hence \(V\) is not a cyclic group. As \(C\) is a cyclic group of order 4, it can only be isomorphic to other cyclic groups of order 4. As \(V\) is not cyclic, then \(C\) and \(V\) are not isomorphic. | M1 (3.1a) | Translates the problem of determining isomorphism by stating that \(V\) is not a cyclic group / \(V\) does not contain a generator / \(V\) only contains self-inverse elements / there is not a one-to-one mapping between the elements that preserves the group operations |
| Concludes that as \(C\) is cyclic and \(V\) is not cyclic then \(C\) is not isomorphic to \(V\), or concludes that as \(C\) contains a generator and \(V\) does not contain a generator then \(C\) is not isomorphic to \(V\), or concludes that as not all elements of \(C\) are self-inverse then \(C\) is not isomorphic to \(V\) | A1 (3.2a) | Any valid conclusion linking the cyclic/non-cyclic distinction or generator argument to non-isomorphism |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| There could be more than one subgroup of \(G\) with order 2. | M1 (2.2b) | Infers that there may not be exactly one subgroup of \(G\) for every factor of the order of \(G\) |
| There is not enough information about \(G\) to determine whether or not it has exactly 5 subgroups, so we cannot tell whether Rachel is correct or incorrect. | A1 (2.3) | Comments on the validity of Rachel's claim and concludes correctly that we cannot be sure about the number of subgroups with orders 2, 4 or 8 |
## Question 9(a)(i):
Translates the problem to that of finding a generator of $C$ | M1 | AO 3.1a
Shows that 1 (or 3) is a generator of $C$ and concludes $C$ is cyclic | A1 | AO 1.1b
Typical solution: If $C$ is a cyclic group, it will contain at least one generator. 1 is a generator of $C$ as repeated addition of 1 modulo 4 will result in all of the elements of $C$. Hence $C$ is a cyclic group.
---
## Question 9(a)(ii):
States the rotations of the square | B1 | AO 2.5
Typical solution: Rotations of the square.
---
## Question 9(b)(i):
Sets up the condition for an element and its inverse: $(-1, 1) \otimes (a, b) = (1, 1)$, so $(-a, b) = (1, 1)$ | M1 | AO 1.1a
Identifies the correct inverse of $(-1, 1)$: Therefore $a = -1$ and $b = 1$, so the inverse of $(-1, 1)$ is $(-1, 1)$ | A1 | AO 1.1b
## Question 9(b)(ii):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $V$ does not contain a generator as all elements in $V$ are self-inverse. Hence $V$ is not a cyclic group. As $C$ is a cyclic group of order 4, it can only be isomorphic to other cyclic groups of order 4. As $V$ is not cyclic, then $C$ and $V$ are not isomorphic. | M1 (3.1a) | Translates the problem of determining isomorphism by stating that $V$ is not a cyclic group / $V$ does not contain a generator / $V$ only contains self-inverse elements / there is not a one-to-one mapping between the elements that preserves the group operations |
| Concludes that as $C$ is cyclic and $V$ is not cyclic then $C$ is not isomorphic to $V$, **or** concludes that as $C$ contains a generator and $V$ does not contain a generator then $C$ is not isomorphic to $V$, **or** concludes that as not all elements of $C$ are self-inverse then $C$ is not isomorphic to $V$ | A1 (3.2a) | Any valid conclusion linking the cyclic/non-cyclic distinction or generator argument to non-isomorphism |
**Subtotal: 2 marks**
---
## Question 9(c):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| There could be more than one subgroup of $G$ with order 2. | M1 (2.2b) | Infers that there may not be exactly one subgroup of $G$ for every factor of the order of $G$ |
| There is not enough information about $G$ to determine whether or not it has exactly 5 subgroups, so we cannot tell whether Rachel is correct or incorrect. | A1 (2.3) | Comments on the validity of Rachel's claim and concludes correctly that we cannot be sure about the number of subgroups with orders 2, 4 or 8 |
**Subtotal: 2 marks**
---
**Question total: 9 marks**
**Question Paper total: 50 marks**9 The group $\left( C , + _ { 4 } \right)$ contains the elements $0,1,2$ and 3
9
\begin{enumerate}[label=(\alph*)]
\item (i) Show that $C$ is a cyclic group.\\
9 (a) (ii) State the group of symmetries of a regular polygon that is isomorphic to $C$\\
9
\item The group ( $V , \otimes$ ) contains the elements (1, 1), (1, -1), (-1, 1) and (-1, -1) The binary operation $\otimes$ between elements of $V$ is defined by
$$( a , b ) \otimes ( c , d ) = ( a \times c , b \times d )$$
9 (b) (i) Find the element in $V$ that is the inverse of $( - 1,1 )$\\
Fully justify your answer.\\[0pt]
[2 marks]\\
9 (b) (ii) Determine, with a reason, whether or not $C \cong V$\\
$\mathbf { 9 }$ (c) The group $G$ has order 16\\
Rachel claims that as $1,2,4,8$ and 16 are the only factors of 16 then, by Lagrange's theorem, the group $G$ will have exactly 5 distinct subgroups, including the trivial subgroup and $G$ itself.
Comment on the validity of Rachel's claim.\\
\includegraphics[max width=\textwidth, alt={}, center]{5ff6e3bb-6392-49cf-b64d-23bc595cd92e-16_2493_1721_214_150}
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Discrete 2023 Q9 [14]}}