51
2.42
Let
G
=
GL
(
2
,
Q
)
, and let
A
=
£
0
−
1
10
±
and
B
=
£
01
−
11
±
. Show that
A
4
=
I
=
B
6
, but that
(
AB
)
n
±=
I
for all
n
>
0. Conclude that
can
have in
f
nite order even though both factors
A
and
B
have
f
nite order.
Solution.
=
·
¸
and
(
)
n
=
·
1
n
¸
.
2.43
(i)
Prove, by induction on
k
≥
1, that
·
cos
θ
−
sin
θ
sin
θ
cos
θ
¸
k
=
·
cos
k
θ
−
sin
k
θ
sin
k
θ
cos
k
θ
¸
.
Solution.
The proof is by induction on
k
. The base step is obvious.
For the inductive step, let
A
=
·
cos
θ
−
sin
θ
sin
θ
cos
θ
¸
.
Then
A
k
+
1
=
AA
k
, and matrix multiplication gives the desired
result if one uses the addition formulas for sine and cosine.
(ii)
Find all the elements of
f
nite order in
SO
(
2
,
R
)
, the special or
thogonal group.
Solution.
By part (i), a matrix
A
=
·
cos
θ
−
sin
θ
sin
θ
cos
θ
¸
.
has
f
nite order if and only if cos
k
α
=
1 and sin
k
α
=
0; that is,
when
k
α
is an integral multiple of 2
π
. Thus,
A
has
f
nite order if
α
=
2
π/
k
for some nonzero integer
k
.
2.44
If
G
is a group in which
x
2
=
1 for every
x
∈
G
, prove that
G
must be
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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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