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79
p

m
!
. By Euclid
’
s lemma, we must have
p

k
for some
k
≤
m
, and this
contradicts
m
<
p
. Therefore,
G
is not simple.
2.135
If
n
≥
3, prove that
A
n
is the only subgroup of
S
n
of order
1
2
n
!
.
Solution.
The case
n
=
3 is very easy, and
n
=
4 has been done in
Exercise 2.89; therefore, we may assume that
n
≥
5. If
H
is a second such
subgroup, then
H
C
S
n
because it has index 2. By the second isomorphism
theorem,
H
∩
A
n
C
A
n
. Since
A
n
is simple, either
H
∩
A
n
=
A
n
or
H
∩
A
n
={
(
1
)
}
. In the
f
rst case,
A
n
≤
H
, so that
A
n
=
H
because they
have the same order. In the second case, we contradict the product formula,
which now says that
1
4
(
n
!
)
2

A
n

H
=
A
n
H
≤
S
n
=
n
!
.
Here is a second proof. If
H
≤
S
n
has index 2, then Proposition 2.97(i)
says that
α
2
∈
H
for every
α
∈
S
n
. Since
α
2
is always an even permutation,
we have
α
2
∈
H
∩
A
n
for all
α
∈
S
n
.How
ev
e
r
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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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