Date | November 2015 | Marks available | 7 | Reference code | 15N.3srg.hl.TZ0.4 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | Prove that | Question number | 4 | Adapted from | N/A |
Question
The binary operation ∗∗ is defined on the set T={0, 2, 3, 4, 5, 6} by a∗b=(a+b−ab)(mod7), a, b∈T.
Copy and complete the following Cayley table for {T, ∗}.
Prove that {T, ∗} forms an Abelian group.
Find the order of each element in T.
Given that {H, ∗} is the subgroup of {T, ∗} of order 2, partition T into the left cosets with respect to H.
Markscheme
Cayley table is
A4
award A4 for all 16 correct, A3 for up to 2 errors, A2 for up to 4 errors, A1 for up to 6 errors
[4 marks]
closed as no other element appears in the Cayley table A1
symmetrical about the leading diagonal so commutative R1
hence it is Abelian
0 is the identity
as x∗0(=0∗x)=x+0−0=x A1
0 and 2 are self inverse, 3 and 5 is an inverse pair, 4 and 6 is an inverse pair A1
Note: Accept “Every row and every column has a 0 so each element has an inverse”.
(a∗b)∗c=(a+b−ab)∗c=a+b−ab+c−(a+b−ab)c M1
=a+b+c−ab−ac−bc+abc A1
a∗(b∗c)=a∗(b+c−bc)=a+b+c−bc−a(b+c−bc) A1
=a+b+c−ab−ac−bc+abc
so (a∗b)∗c=a∗(b∗c) and ∗ is associative
Note: Inclusion of mod 7 may be included at any stage.
[7 marks]
0 has order 1 and 2 has order 2 A1
32=4, 33=2, 34=6, 35=5, 36=0 so 3 has order 6 A1
42=6, 43=0 so 4 has order 3 A1
5 has order 6 and 6 has order 3 A1
[4 marks]
H={0, 2} A1
0∗{0, 2}={0, 2}, 2∗{0, 2}={2, 0}, 3∗{0, 2}={3, 6}, 4∗{0, 2}={4, 5},
5∗{0, 2}={5, 4}, 6∗{0, 2}={6, 3} M1
Note: Award the M1 if sufficient examples are used to find at least two of the cosets.
so the left cosets are {0, 2}, {3, 6}, {4, 5} A1
[3 marks]
Total [18 marks]
Examiners report