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Past exam of the mathematics course of the University of Cambridge
/
2021
/
ia
/
Paper 3
/
1D
/
i
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Mathematics course of the University of Cambridge
Past exam of the mathematics course of the University of Cambridge
2021
ia
Paper 3
1D
Created
2026-09-24
Updated
2026-09-24
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Solution
i
Solution
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i
Suppose
G
/
Z
(
G
)
is generated by
g
Z
(
G
)
. Every element of
G
can then be written
g
a
z
with
z
∈
Z
(
G
)
. Hence
(
g
a
z
1
)
(
g
b
z
2
)
=
g
a
+
b
z
1
z
2
=
(
g
b
z
2
)
(
g
a
z
1
)
.
(1)
Thus the
cyclic quotient by the center
condition
forces
G
to be abelian.
If
∣
H
/
Z
(
H
)
∣
=
7
, the
quotient
has prime order and is therefore cyclic. The result just proved would make
H
abelian, so
Z
(
H
)
=
H
and the
quotient
would have order one,
a
contradiction. Therefore
no such group
H
exists
.
(2)
Ancestors
(10)
1D
Paper 3
ia
2021
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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