Edexcel FP2 2019 June — Question 6 12 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2019
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeProve group-theoretic identities
DifficultyChallenging +1.2 This is a multi-part group theory question requiring proof of associativity (algebraic manipulation), demonstrating a group is cyclic (finding a generator), listing subgroups (systematic enumeration), and describing an isomorphism (pattern recognition). While covering several concepts, each part follows standard procedures taught in Further Maths without requiring deep insight or novel problem-solving approaches. The associativity proof is routine algebra, and the finite group analysis involves straightforward computation with small sets.
Spec8.03a Binary operations: and their properties on given sets8.03f Subgroups: definition and tests for proper subgroups8.03g Cyclic groups: meaning of the term8.03l Isomorphism: determine using informal methods

    1. A binary operation * is defined on positive real numbers by
$$a * b = a + b + a b$$ Prove that the operation * is associative.
(ii) The set \(G = \{ 1,2,3,4,5,6 \}\) forms a group under the operation of multiplication modulo 7
  1. Show that \(G\) is cyclic. The set \(H = \{ 1,5,7,11,13,17 \}\) forms a group under the operation of multiplication modulo 18
  2. List all the subgroups of \(H\).
  3. Describe an isomorphism between \(G\) and \(H\).

Question 6(i):
AnswerMarks Guidance
WorkingMark Guidance
\((a*b)*c = (a+b+ab)*c = a+b+ab+c+(a+b+ab)c\)M1 2.1
\(a*(b*c) = a*(b+c+bc) = a+b+c+bc+a(b+c+bc)\)M1 2.1
\(a+b+ab+c+(a+b+ab)c = a+b+c+bc+ab+ac+abc\) \(= a+b+c+bc+a(b+c+bc)\)A1 2.2a
So \((a*b)*c = a*(b*c)\), which means \(*\) is associativeA1 2.4
Notes:
- M1: Begins proof by correctly expanding \((a*b)*c\) or \(a*(b*c)\) to an expression in \(a\), \(b\), \(c\)
- M1: Makes progress by attempting to expand both \((a*b)*c\) and \(a*(b*c)\)
- A1: Both underlined expressions correct with correct expansion seen for each independently
- A1: Explains \((a*b)*c = a*(b*c)\) means \(*\) is associative; depends on both M marks
Question 6(ii)(a):
AnswerMarks Guidance
WorkingMark Guidance
\(3^2=2,\ 3^3=6,\ 3^4=4,\ 3^5=5,\ 3^6=1\) or \(5^2=4,\ 5^3=6,\ 5^4=2,\ 5^5=3,\ 5^6=1\)M1 2.1
3 (or 5) has order 6 and so generates the group so \(G\) is cyclicA1 2.4
Notes:
- M1: Demonstrates understanding of cyclic by attempting all powers of 3 or 5; accept \(\langle 3\rangle = \{3,2,6,4,5,1\}\)
- A1: Must evaluate all powers correctly and explain why \(G\) is cyclic; must refer to cyclic in conclusion
- Special case: Allow M1A0 if order of 3 (or 5) stated as 6 without justification but must reference generator or same order as \(G\)
Question 6(b):
AnswerMarks Guidance
WorkingMark Guidance
\(\{1\}\), \(H\)B1 1.1b
\(\{1,17\}\) or \(\{1,7,13\}\)M1 1.1b
\(\{1,17\}\) and \(\{1,7,13\}\) (and no others)A1 1.1b
Notes:
- B1: Identifies \(\{1\}\) and \(H\) as subgroups
- M1: Identifies \(\{1,17\}\) or \(\{1,7,13\}\) as a subgroup
- A1: Identifies \(\{1,17\}\) and \(\{1,7,13\}\) as subgroups and no others
Question 6(c):
AnswerMarks Guidance
WorkingMark Guidance
\(\begin{array}{c\cccccc} G & 1 & 2 & 3 & 4 & 5 & 6 \\ H & 1 & 7 & 5 & 13 & 11 & 17 \end{array}\) or \(\begin{array}{c\ cccccc} G & 1 & 2 & 3 & 4 & 5 & 6 \\ H & 1 & 13 & 11 & 7 & 5 & 17 \end{array}\)
Notes:
- M1: Attempts to identify an isomorphism — at least 2 correct non-identity pairings, or rearranges group tables, or maps powers of a generator e.g. \((\text{their } 3)^k \to (\text{their } 5)^k\), or matches non-trivial proper subgroups
- A1: Identifies 4 correct pairings, or sets up mapping with one correct generator
- A1: All pairings correct, or sets up mapping with generators of each group correct, e.g. \(3^k \to 5^k\)
# Question 6(i):

| Working | Mark | Guidance |
|---------|------|----------|
| $(a*b)*c = (a+b+ab)*c = a+b+ab+c+(a+b+ab)c$ | M1 | 2.1 |
| $a*(b*c) = a*(b+c+bc) = a+b+c+bc+a(b+c+bc)$ | M1 | 2.1 |
| $a+b+ab+c+(a+b+ab)c = a+b+c+bc+ab+ac+abc$ $= a+b+c+bc+a(b+c+bc)$ | A1 | 2.2a |
| So $(a*b)*c = a*(b*c)$, which means $*$ is associative | A1 | 2.4 |

**Notes:**
- M1: Begins proof by correctly expanding $(a*b)*c$ **or** $a*(b*c)$ to an expression in $a$, $b$, $c$
- M1: Makes progress by attempting to expand both $(a*b)*c$ **and** $a*(b*c)$
- A1: Both underlined expressions correct with correct expansion seen for each independently
- A1: Explains $(a*b)*c = a*(b*c)$ means $*$ is associative; depends on both M marks

---

# Question 6(ii)(a):

| Working | Mark | Guidance |
|---------|------|----------|
| $3^2=2,\ 3^3=6,\ 3^4=4,\ 3^5=5,\ 3^6=1$ **or** $5^2=4,\ 5^3=6,\ 5^4=2,\ 5^5=3,\ 5^6=1$ | M1 | 2.1 |
| 3 (or 5) has order 6 and so generates the group so $G$ is cyclic | A1 | 2.4 |

**Notes:**
- M1: Demonstrates understanding of cyclic by attempting all powers of 3 or 5; accept $\langle 3\rangle = \{3,2,6,4,5,1\}$
- A1: Must evaluate all powers correctly and explain why $G$ is cyclic; must refer to cyclic in conclusion
- Special case: Allow M1A0 if order of 3 (or 5) stated as 6 without justification but must reference generator or same order as $G$

---

# Question 6(b):

| Working | Mark | Guidance |
|---------|------|----------|
| $\{1\}$, $H$ | B1 | 1.1b |
| $\{1,17\}$ **or** $\{1,7,13\}$ | M1 | 1.1b |
| $\{1,17\}$ **and** $\{1,7,13\}$ (and no others) | A1 | 1.1b |

**Notes:**
- B1: Identifies $\{1\}$ and $H$ as subgroups
- M1: Identifies $\{1,17\}$ **or** $\{1,7,13\}$ as a subgroup
- A1: Identifies $\{1,17\}$ **and** $\{1,7,13\}$ as subgroups and no others

---

# Question 6(c):

| Working | Mark | Guidance |
|---------|------|----------|
| $\begin{array}{c\|cccccc} G & 1 & 2 & 3 & 4 & 5 & 6 \\ H & 1 & 7 & 5 & 13 & 11 & 17 \end{array}$ **or** $\begin{array}{c\|cccccc} G & 1 & 2 & 3 & 4 & 5 & 6 \\ H & 1 & 13 & 11 & 7 & 5 & 17 \end{array}$ | M1, A1, A1 | 3.1a, 1.1b, 1.1b |

**Notes:**
- M1: Attempts to identify an isomorphism — at least 2 correct non-identity pairings, or rearranges group tables, or maps powers of a generator e.g. $(\text{their } 3)^k \to (\text{their } 5)^k$, or matches non-trivial proper subgroups
- A1: Identifies 4 correct pairings, or sets up mapping with one correct generator
- A1: All pairings correct, or sets up mapping with generators of each group correct, e.g. $3^k \to 5^k$

---
\begin{enumerate}
  \item (i) A binary operation * is defined on positive real numbers by
\end{enumerate}

$$a * b = a + b + a b$$

Prove that the operation * is associative.\\
(ii) The set $G = \{ 1,2,3,4,5,6 \}$ forms a group under the operation of multiplication modulo 7\\
(a) Show that $G$ is cyclic.

The set $H = \{ 1,5,7,11,13,17 \}$ forms a group under the operation of multiplication modulo 18\\
(b) List all the subgroups of $H$.\\
(c) Describe an isomorphism between $G$ and $H$.

\hfill \mbox{\textit{Edexcel FP2 2019 Q6 [12]}}