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Past exam of the mathematics course of the University of Cambridge
/
2019
/
iii
/
Paper 138
/
4
/
b
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Mathematics course of the University of Cambridge
Past exam of the mathematics course of the University of Cambridge
2019
iii
Paper 138
4
2026-10-03
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Table of contents
i
b
Solution
i
ii
b
Solution
ii
i
0
0
0
b
Solution
0
0
0
i
Set
α
=
[
G
:
H
]
−
1
id
M
∈
End
R
H
(
M
)
.
(1)
Every conjugate of
α
equals
α
, so
Tr
H
G
(
α
)
=
[
G
:
H
]
α
=
id
M
.
(2)
The
D. Higman criterion
therefore gives
every
RG
-module is relatively
H
-projective
.
(3)
ii
0
0
0
b
Solution
0
0
0
ii
If
M
is projective over
RG
, then its restriction is projective over
R
H
because
RG
is
a
free
right
R
H
-
module
and
a
free
RG
-
module
restricts to
a
free
R
H
-
module
.
Conversely, suppose
Res
H
G
M
is projective. Then
Ind
H
G
Res
H
G
M
(1)
is projective over
RG
. Part (
i
) says that
M
is
a
direct summand of this induced
module
, so
M
is projective. Hence
projectivity detected on a subgroup of invertible index
gives
M
is
RG
-projective
⟺
Res
H
G
M
is
R
H
-projective
.
(2)
Ancestors
(10)
4
Paper 138
iii
2019
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
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