Question 2:
2 | (a) | From Lagrange’s Theorem the order of any
subgroup of G must be a factor of 8 and 6 is not
a factor of 8 | B1
[1] | 2.4 | Or “order of any subgroup of G
(or an group of order 8) must be 1,
2 or 4 (or 8)” or “order of any
subgroup must be a factor of the
order of the group and 6 is not a
factor of 8”. | If referenced, Lagrange’s Theorem
does not have to be quoted provided
that it is applied. So B1 for eg “6 is
not a factor of 8 so by Lagrange’s
Theorem there can be no subgroup
of G of order 6” but B0 for eg “By
Lagrange’s Theorem there can be
no subgroup of G of order 6”.
2 | (b) | g2 (or g6)
g6 (or g2) and no other | B1
B1
[2] | 2.2a
2.2a | May see eg gg or g◦g used here
and/or throughout. Allow any
multiplicative notation and any
symbol for a binary operation.
2 | (c) | e↔0
g↔1, g2↔2, g3↔3, g4↔4, g5↔5, g6↔6, g7↔7
g↔3, g2↔6, g3↔1, g5↔7, g6↔2, g7↔5
g↔5, g2↔2, g3↔7, g5↔1, g6↔6, g7↔3 and
g↔7, g2↔6, g3↔5, g5↔3, g6↔2, g7↔1 | B1
B1
B1
B1 | 2.2a
2.2a
2.2a
2.2a | Only needs to be seen once.
Any one.
Any other.
Other two. Ignore repeats. | g4↔4 does need not be seen again
g4↔4 does need not be seen again
Alternative method:
e↔0 | B1 | Only needs to be seen once
Either g↔1 or g↔3 or g↔5 or g↔7 | M1 | M1 | Giving all 4 possible isomorphism
options for any generator of G (ie
g, g3, g5 or g7)
g↔1, g2↔2, g3↔3, g4↔4, g5↔5, g6↔6, g7↔7 | A1 | Completing the specification of
any one isomorphism
g↔3, g2↔6, g3↔1, g5↔7, g6↔2, g7↔5 and | A1 | Other three. Ignore repeats. | Other three. Ignore repeats. | g4↔4 does need not be seen again
g↔5, g2↔2, g3↔7, g5↔1, g6↔6, g7↔3 and
g↔7, g2↔6, g3↔5, g5↔3, g6↔2, g7↔1
[4]