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1.10. GROUPS
69
Moreover, if
p
is odd and
k
is not divisible by
p
, then
.1
C
kp/
p
s
6±
1 .
mod
p
s
C
2
/:
Hint
: Use induction on s.
1.9.11.
(a)
Suppose that the integer
a
is relatively prime to the prime
p
.
Then for all integers
s
²
0
,
a
p
s
.p
±
1/
±
1 .
mod
p
s
C
1
/:
Hint
: By Fermat’s little theorem,
a
p
±
1
D
1
C
kp
for some in
teger
k
. Thus
a
p
s
.p
±
1/
D
.1
C
kp/
p
s
. Now use the previous
exercise.
(b)
Show that the statement in part (a) is the special case of Euler’s
theorem, for
n
a power of a prime.
1.9.12.
Let
n
be a natural number with prime factorization
n
D
p
k
1
1
³³³
p
k
s
s
,
and let
a
be an integer that is relatively prime to
n
.
(a)
Fix an index
i
(
1
´
i
´
s
) and put
b
D
a
Q
j
¤
i
'.p
k
j
j
/
. Show
b
is also relatively prime to
n
.
(b)
Show that
a
'.n/
D
b
'.p
k
i
i
/
, and apply the previous exercise to
show that
a
'.n/
±
1 .
mod
p
k
i
i
/
.
(c)
Observe that
a
'.n/
µ
1
is divisible by
p
k
i
i
for each
i
. Conclude
that
a
'.n/
µ
1
is divisible by
n
. This is the conclusion of Euler’s
theorem.
1.10. Groups
An
operation
or
product
on a set
G
is a function from
G
¶
G
to
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 Fall '08
 EVERAGE
 Algebra, Integers

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