AS June 2024 Q1

EdexcelCurrent spec9 marksGroups

(i) The table below is a Cayley table for the group \(G\) with operation \(\circ\)

\(\circ\)	\(a\)	\(b\)	\(c\)	\(d\)	\(e\)	\(f\)
\(a\)	\(d\)	\(c\)	\(b\)	\(a\)	\(f\)	\(e\)
\(b\)	\(e\)	\(f\)	\(a\)	\(b\)	\(c\)	\(d\)
\(c\)	\(f\)	\(e\)	\(d\)	\(c\)	\(b\)	\(a\)
\(d\)	\(a\)	\(b\)	\(c\)	\(d\)	\(e\)	\(f\)
\(e\)	\(b\)	\(a\)	\(f\)	\(e\)	\(d\)	\(c\)
\(f\)	\(c\)	\(d\)	\(e\)	\(f\)	\(a\)	\(b\)

(a) State which element is the identity of the group. (1)

(b) Determine the inverse of the element \((b \circ c)\) (2)

(d) Show that the set \(\{b, d, f\}\) is a subgroup of \(G\). (2)

(ii) Given that \(H\) is a group with an element \(x\) of order 3 and an element \(y\) of order 6 satisfying\[yx = xy^5\]show that \(y^3xy^3x^2\) is the identity element. (3)

Scheme	Marks	AO
Identity is \(d\)	B1	2.2a
	(1)

Mark scheme (i)(b)

Scheme	Marks	AO
\((b \circ c)^{-1} = a^{-1}\) Alt: \((b \circ c)^{-1} = c^{-1} \circ b^{-1} = c \circ f\)	M1	1.1a
\(= a\) Alt: \(= a\)	A1	1.1b
	(2)

Notes

M1: Correct method to find the inverse of \(b \circ c\). May find \(b \circ c\) first and identify its inverse, or may apply inverse property and find \(c \circ f\).

A1: Correct element, \(a\).

Mark scheme (i)(c)

Scheme	Marks	AO
The set cannot be a subgroup because e.g. \(G\) has order 6 and 4 does not divide 6 so by Lagrange’s Theorem, or the identity is not in the set, or closure fails as e.g. \(a \circ a = d\) is not in the set.	B1	2.4
	(1)

Notes

B1: Any correct reason given. See scheme for possibilities.

Mark scheme (i)(d)

Scheme

Marks

Table for \(\{b, d, f\}\) is

	\(b\)	\(d\)	\(f\)
\(b\)	\(f\)	\(b\)	\(d\)
\(d\)	\(b\)	\(d\)	\(f\)
\(f\)	\(d\)	\(f\)	\(b\)

2.1

This is closed, as no new elements, \(b\) and \(f\) are inverses and \(d\) is the identity, so the subset is a subgroup.

1.1b

(2)

Notes

M1: Investigates closure, e.g. by drawing the table for \(\{b, d, f\}\) or finding the individual products (at least 3) or by considering \(\langle b \rangle = \{b, b^2, b^3\} = \{b, f, d\}\) so cyclic.

A1: Correctly shows closure, and refers to inverse and identity, and makes conclusion it is a subgroup.

Mark scheme (ii)

Scheme	Marks	AO
\(y^3xy^3x^2 = y^2yxy^3x^2 = y^2xy^5y^3x^2\)	M1	1.1b
\(= yyxy^8x^2 = yxy^5y^8x^2 = xy^5y^{13}x^2\)	M1	3.1a
\(= xy^{18}x^2 = xex^2 = x^3 = e\) *	A1*	2.1
	(3)
	(9 marks)

Notes

M1: Applies the relation \(yx = xy^5\) at least once to change positions of an \(x\) and a \(y\). May work from either end.

M1: Completes the process of gathering all \(y\)’s or \(x\)’s together by using the relation repeatedly. May reduce elements according to order throughout rather than at the end.

A1*: Uses the orders of the elements to reduce gathered \(y\) terms and \(x\) terms down to identity to complete the proof.