10
are
roots
of
x
10
−
1,
and
the
elements
of
order
15
are
roots
of
x
15
−
1.
Unfortunately,
6
+
10
+
15
=
31.
This
is
too
large
to
draw
a
conclusion
about
elements
of
order
30.
The
problem
is
caused
by
double
counting.
For
example,
the
elements
of
order
3
are
roots,
both
of
x
6
−
1
and
of
x
15
−
1.
When
one
eliminates
the
double
counting,
one
sees
that
there
must
be
elements
of
order
30.
�
It
is
fussy
arithmetic
to
make
a
proof
based
on
the
method
illustrated
by
these
examples.
We
use
a
lemma
about
the
orders
of
elements
of
an
abelian
group.
Lemma
2.
(a)
Let
u
and
v
be
elements
of
an
abelian
group
G
,
of
finite
orders
a
and
b
,
respectively,
and
let
m
be
the
least
common
multiple
of
a
and
b
.
Then
G
contains
an
element
of
order
m
.
(b)
Let
G
be
a
finite
abelian
group,
and
let
m
be
the
least
common
multiple
of
the
orders
of
elements
of
G
.
Then
G
contains
an
element
of
order
m
.
Note:
The
hypothesis
that
G
be
abelian
is
essential
here.
The
symmetric
group
S
3
,
which
is
not
abelian,
contains
elements
of
orders
2
and
3
but
no
element
of
order
6.
proof
of
the
theorem.
We’ll
prove
the
theorem,
assuming
that
Lemma
2
has
been
proved.
Let
m
be
the
least
common
multiple
of
the
orders
of
the
elements
of
F
×
p
.
The
lemma
tells
us
that
F
×
p
contains
an
element
�
of
order
m
.
Therefore
m
divides
the
order
of
the
group,
which
is
p
−
1,
and
m
�
p
−
1.
Also,
since
m
is
the
least
common
multiple
of
the
orders
of
the
elements
of
F
×
p
,
the
order
of
every
element
divides
m
.
So
every
element
of
F
×
p
is
a
root
of
the
polynomial
x
m
−
1.
Since
this
polynomial
has
at
most
m
roots,
p
−
1
�
m
.

#### You've reached the end of your free preview.

Want to read all 3 pages?

- Fall '13
- Prime number, Greatest common divisor, Divisor, Abelian group