64
Solution.
It is easy to see that
ϕ
is a homomorphism:
ϕ(
MN
)
=
QMNQ
−
1
=
QMQ
−
1
QNQ
−
1
=
M
)ϕ(
N
).
Next, if
M
=
·
ab
cd
¸
,
then
a
+
c
=
1 and
b
+
d
=
1. It follows that im
ϕ
≤
A
, for matrix
multiplication gives
·
10
11
¸·
−
¸
=
·
a
−
bb
01
¸
(the bottom row of the last matrix is
(
a
+
c
)
−
(
b
+
d
)
b
+
d
,
and this is 0 1 because
M
is stochastic.
Finally, we must show that
ϕ
is surjective. This is obvious from
the last calculation. If
A
=
·
¸
∈
A
,
then
A
=
M
)
, where
M
=
·
a
+
1
−
a
−
b
1
−
b
¸
.
2.91
Prove that the symmetry group
6(π
n
)
, where
π
n
is a regular polygon with
n
vertices, is isomorphic to a subgroup of
S
n
.
Solution.
Every isometry
ϕ
∈
n
)
permutes the
n
vertices
X
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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