1.2.
Semigroups,
Monoids,
and
Groups
7
is
called
a
partition
of
G by H. We
write
H G if H is
normal
in G. If H is a
normal
subgroup, then
the
partition
of G by H is
compatible. Using normality
and H
2
= H, we
have
On
the
other hand, every compatible partition
of G can be
given
by a
normal subgroup.
Then
the set G/H = (gH \ g G}
forms
a
factor
group
of G
having
Hg
1
Hg
2
=
Hg
1
g2,
Hg
1
,
Hg
2
G/H, where
H (= 1H) is the
identity element
of G/H and
(Hg)
-1
=
Hg
-l
for
each
Hg
G/H. However,
if : G T is a
surjective homomorphism, then
the
kernel
of , ker = K = {g G : (g) =
(l)},
is
easily seen
to be a
normal subgroup,
and, moreover,
G/K is
isomorphic
to T via the map gK
(gK)
=
(g). Thus
G
maps
homomorphically onto
G/H
when
H is
normal
and
every homomorphic image
of G is of
this form.
Obviously,
G and {1} are
normal subgroups
of G. If N is a
normal subgroup
of G
such
that
{1} N G,
then
it is
called
a
proper normal
subgroup
ofG,
and
then
G/N is
said
to be
a
proper
factor
group
ofG.
If G is
nontrivial
and has no
proper normal subgroups,
then
it is
called
simple.
If G is
simple, then every homomorphic image
of G is
isomorphic
to {1} or G. A
proper normal subgroup
N G is
maximal
if N H G
implies
H = N
or
H = G.
This holds
if and
only
if G/N is
simple.
In
an
abelian group, every subgroup
is
normal.
A
group
is
cyclic
if it can be
generated
as a
group
by one of its
elements.
An
elementary exercise shows that every cyclic group
is
necessarily abelian,
but not
conversely.
If an
abelian group
G is
simple,
it
contains
an
element
g
1, and
since
g is
normal
in G, we
have that
g = G.
Thus
G is
cyclic.
If
o(g)
= ,
G
must
be
isomorphic
to the
group
Z of
integers under addition,
but
this
is not
simple since
the
even integers
form
a
(normal) subgroup. Hence, o(g)
, but
then
the
order
of g
must
be
prime:
for if
o(g)
= nm, for
integers
n, m > 0,
then
g = G has nm
elements.
Suppose
m 1;
then
by
definition
of
order
of g, g
n
1 and so the
normal subgroup
g
n
must
be G. Now
since (g
n
)
m
= g
nm
= 1, the
group
g
n
=
{g
n
,...,
(g
n
)
m
}
= G has no
more than
m
elements.
So it can
only
be
that
n = 1,
whence o(g)
is
prime. Thus
an
abelian
group
is
simple
if and
only
if it is a finite
cyclic group
of
prime order.
It is
immediate that
for
every simple
(finite
or
infinite)
group
G, G is
nonabelian
if and
only
if G is not
cyclic.
If
any two
semigroups
S
1
and S
2
are
considered,
the
ordered pairs (s
1
,
s
2
)
with
s
1
from
S
1
and s
2
from s
2
form
a
semigroup according
to the
rule that
defines
products
of
pairs componentwise, i.e.,
(s ,s )(s , s ) = (s s , s s ), s ,s
S
1
,s
,s S
2
.
This
semigroup
is
called
the
direct product
ofS
1
and 52 and is
written
S
1
x S
2
. It is
easy
to see
that
S
1
x S
2
and S
2
x S
1
are
isomorphic.
If e
2
is an
idempotent
in S
2
,
then
the
ordered pairs
(s
1
,
e
2
)
with
s
1
S
1
make
up a
subsemigroup
of S
1
x S
2
isomorphic
to S
1
. In
particular,
if
S
1
has an
identity
e
1
and S
2
has an
identity
e
2
,
then
(e
1
,
e
2
) is an
identity
for S
1
x S
2
.
Thus
the
direct
product
of
monoids
is
obviously
a
monoid,
and the
component monoids
are
isomorphic
to
submonoids
of the
direct product.
The
analogous assertions hold
for
groups since inverses
in the
direct product
of
groups
can be
obtained
by
taking inverses
in
each component.
Next
we
prove
the
following theorem.
Theorem
1.2.
Let G =
{g=,...,
g
n
} be a
(finite)
order
n
group.
Put P
G
=
{g
P(1)
. . .
g
P(n)
P is a
permutation over
{1,...,n}}.IfGis
simple
and
noncommutative, then
there
exists
a
positive
integer
m
with
P = G.