Question 7:
--- 7(a) ---
7(a) | Deduces correctly that G has
two (distinct) subgroups
PI by reference to subgroups of
order 1 and p | 2.2a | M1 | As p is prime, it has exactly two
factors: 1 and p. Hence by
Lagrange’s theorem, G has two
subgroups.
Completes rigorous
mathematical argument with
reference to 1 and p as the
factors of the prime number p
or 1 and p as the orders of the
two subgroups
and
by naming or explaining
Lagrange’s theorem | 2.1 | R1
Total | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b)(i) ---
7(b)(i) | Uses correct mathematical
language of ‘generator’ to name
the element g | 2.5 | B1 | Generator
Total | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b)(ii) ---
7(b)(ii) | States correctly that G is
isomorphic to the cyclic group of
order p (or a specific example of
a cyclic group of order p) | 1.1b | B1 | Cyclic group of order p
Total | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c) ---
7(c) | Uses the definition of an inverse
element to set up a relationship
r
of the form g x=e
r
or x  g = e
PI | 1.1a | M1 | k r
L e g t b e th e in v e r s e o g f
r k p
g  g = e = g
r + k p
g = g  k = p − r
r p − r
T h e in v e r s e o f g is g
p−r
Finds correctly g , the
r
inverse of g | 1.1b | A1
Total | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(d)(i) ---
7(d)(i) | Explains correctly that addition
modulo 5 between two elements
of G will only ever result in 0, 1,
2, 3 or 4 which are all in G | 2.4 | E1 | For any a and b in G, a + b (mod 5)
is either 0, 1, 2, 3 or 4, all of which
are in G.
Total | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 7(d)(ii) ---
7(d)(ii) | Completes the Cayley table
correctly | 1.1b | B1 | 0 1 2 3 4

0 0 1 2 3 4
1 1 2 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3
Total | 1
0 | 1 | 2 | 3 | 4