Math 5125
Friday, September 30
September 30, Ungraded Homework
Exercise 6.3.5 on page 220
Establish a finite presentation for
A
4
using 2 generators.
Let
F
denote the free group on
{
x
,
y
}
, let
a
= (
1 2
)(
3 4
)
,
b
= (
1 2 3
)
, and define
θ
:
F
→
A
4
by
θ
x
=
a
,
θ
y
=
b
. Let
K
=
ker
θ
. It is easy to check that
a
,
b
=
A
4
. Indeed by Lagrange’s
theorem 6

a
,
b

12, so if
a
,
b
=
A
4
, then

a
,
b

=
6. But a group of order 6 has a unique
subgroup of order 3, so has exactly 2 elements of order 3. Since
b
,
b
2
,
aba
are three distinct
elements of order 3, we have a contradiction. Thus
a
,
b
=
A
4
as asserted and we conclude
that
θ
is onto. Therefore
F
/
K
∼
=
A
4
by the fundamental homomorphism theorem.
Next note that
θ
x
2
=
θ
y
3
=
θ
(
xy
)
3
=
1. In view of this, we will try
x
2
,
y
3
,
(
xy
)
3
for our set
of relators. Let
N
denote the normal closure of
{
x
2
,
y
3
,
(
xy
)
3
}
in
F
. Then
x
,
y

x
2
=
1
,
y
3
=
1
,
(
xy
)
3
=
1
is by definition isomorphic to
F
/
N
. We want to show that
F
/
N
∼
=
A
4
, that is
F
/
N
∼
=
F
/
K
. Since
x
2
,
y
3
,
(
xy
)
3
∈
K
, we certainly have
N
⊆
K
. We want to show equality,
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 Fall '07
 PALinnell
 Math, Algebra, kernel, Fundamental theorem on homomorphisms, Group homomorphism, xa yb xc

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