ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE
GROUPS
PETE L. CLARK
1.
Reminder on quotient groups
Let
G
be a group and
H
a subgroup of
G
. We have seen that the left cosets
xH
of
H
in
G
give a partition of
G
.
Motivated by the case of quotients of rings by
ideals, it is natural to consider the product operation on cosets. Recall that for any
subsets
S, T
of
G
, by
ST
we mean
{
st

s
∈
S, t
∈
T
}
.
If
G
is commutative, the product of two left cosets is another left coset:
(
xH
)(
yH
) =
xyHH
=
xyH.
In fact, what we really used was that for all
y
∈
G
,
yH
=
Hy
. For an arbitrary
group
G
, this is a property of the subgroup
H
, called
normality
. But it is clear –
and will be good enough for us – that if
G
is commutative, all subgroups are normal.
If
G
is a group and
H
is a normal subgroup, then the set of left cosets, denoted
G/H
, itself forms a group under the above product operation, called the
quotient
group
of
G
by
H
. The map which assigns
x
∈
G
to its coset
xH
∈
G/H
is in fact
a surjective group homomorphism
q
:
G
→
G/H
, called the
quotient map
(or in
common jargon, the “natural map”), and its kernel is precisely the subgroup
H
.
Theorem 1.
(Isomorphism theorem) Let
f
:
G
→
G
′
be a surjective homomor
phism of groups, with kernel
K
. Then
G/K
is isomorphic to
G
′
.
Proof.
We define the isomorphism
q
(
f
) :
G/K
→
G
′
in terms of
f
: map the coset
xK
to
f
(
x
)
∈
G
′
. This is welldefined, because if
xK
=
x
′
K
, then
x
′
=
xk
for some
k
∈
K
, and then
f
(
x
′
) =
f
(
x
)
f
(
k
) =
f
(
x
)
·
e
=
f
(
x
)
,
since
k
is in the kernel of
f
. It is immediate to check that
q
(
f
) is a homomorphism
of groups. Because
f
is surjective, for
y
∈
G
′
there exists
x
∈
G
such that
f
(
x
) =
y
and then
q
(
f
)(
xK
) =
y
, so
q
(
f
) is surjective.
Finally, if
q
(
f
)(
xK
) =
e
, then
f
(
x
) =
e
and
x
∈
K
, so
xK
=
K
is the identity element of
G/K
.
In other words, a group
G
′
is (isomorphic to) a quotient of a group
G
iff there
exists a surjective group homomorphism from
G
to
G
′
.
Corollary 2.
If
G
and
G
′
are finite groups such that there exists a surjective group
homomorphism
f
:
G
→
G
′
, then
#
G
′

#
G
.
Proof.
G
′
∼
=
G/
ker
f
, so #
G
′
·
#(ker
f
) = #
G
.
1
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PETE L. CLARK
Remark: Suitably interepreted, this remains true for infinite groups.
Corollary 3.
(“transitivity of quotients”) If
G
′
is isomorphic to a quotient group
of
G
and
G
′′
is isomorphic to a quotient group of
G
′
, then
G
′′
is isomorphic to a
quotient group of
G
.
Proof.
We have surjective group homomorphisms
q
1
:
G
→
G
′
and
q
2
:
G
′
→
G
′′
,
so the composition
q
2
◦
q
1
is a surjective group homomorphism from
G
to
G
′′
.
2.
Cyclic groups
Recall that a group
G
is cyclic if there exists some element
g
in
G
such that every
x
in
g
is of the form
g
n
for some integer
n
. (Here we are using the conventions that
g
0
=
e
is the identity element of
G
and that
g
−
n
= (
g
−
1
)
n
.) Such an element
g
is
called a generator. In general, a cyclic group will have more than one generator,
and it is a numbertheoretic problem to determine how many generators there are.
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 Spring '11
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 Algebra, Number Theory, Sets, Abelian group, pete l. clark

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