77
(iv)
Use parts (ii) and (iii) to give a second proof of the simplicity
of
A
5
.
Solution.
Since
H
contains 1, the order of
H
is a sum of 1 together
with some of the numbers 12, 12, 15, and 20. The only such sum
that divides 60 is 60 itself.
2.126
If
σ,τ
∈
S
5
, where
σ
is a 5cycle and
τ
is a transposition, prove that
h
σ,τ
i=
S
5
.
Solution.
Absent.
2.127
(i)
For all
n
≥
3, prove that every
α
∈
A
n
is a product of 3cycles.
Solution.
We show that
(
123
)
and
(
ijk
)
are conjugate in
A
n
(and thus that all 3cycles are conjugate in
A
n
). If these cy
cles are not disjoint, then each
f
xes all the symbols outside of
{
1
,
2
,
3
,
i
,
j
]}
, say, and the two 3cycles lie in
A
∗
, the group of all
even permutations on these 5 symbols. Of course,
A
∗
∼
=
A
5
, and,
as in Lemma 2.155,
(
123
)
and
(
ijk
)
are conjugate in
A
∗
;
a for
tiori
, they are conjugate in
A
n
. If the cycles are disjoint, then we
have just seen that
(
123
)
is conjugate to
(
3
jk
)
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 Fall '11
 KeithCornell
 Group Theory, Normal subgroup, σ, Subgroup, A∗

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