Difference between revisions of "Normal subgroup"
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A '''normal subgroup''' <math>{\rm H}</math> of a [[group]] <math>{\rm G}</math> is a [[subgroup]] of <math>{\rm G}</math> for which the relation "<math>xy^{-1} \in {\rm H}</math>" of <math>x</math> and <math>y</math> is compatible with the law of composition on <math>{\rm G}</math>, which in this article is written multiplicatively. The [[quotient group]] of <math>{\rm G}</math> under this relation is often denoted <math>{\rm G/H}</math> (said, "<math>{\rm G}</math> mod <math>{\rm H}</math>"). (Hence the notation <math>\mathbb{Z}/n\mathbb{Z}</math> for the integers mod <math>n</math>.) | A '''normal subgroup''' <math>{\rm H}</math> of a [[group]] <math>{\rm G}</math> is a [[subgroup]] of <math>{\rm G}</math> for which the relation "<math>xy^{-1} \in {\rm H}</math>" of <math>x</math> and <math>y</math> is compatible with the law of composition on <math>{\rm G}</math>, which in this article is written multiplicatively. The [[quotient group]] of <math>{\rm G}</math> under this relation is often denoted <math>{\rm G/H}</math> (said, "<math>{\rm G}</math> mod <math>{\rm H}</math>"). (Hence the notation <math>\mathbb{Z}/n\mathbb{Z}</math> for the integers mod <math>n</math>.) | ||
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− | <math>N</math> is said to be a normal subgroup of a group <math>G</math> if <math>aNa^{-1}=N</math>.Note that this means <math>aN=Na</math> but it does not imply that for every <math>n\in N</math>, <math>an=na</math>. | + | An equivalent definition of normal subgroups is this- |
+ | |||
+ | <math>N</math> is said to be a normal subgroup of a group <math>G</math> if <math>aNa^{-1}=N</math>.Note that this means <math>aN=Na</math> but it does not imply that for every <math>n\in N</math>, <math>an=na</math>. | ||
== Description == | == Description == |
Revision as of 01:32, 22 December 2012
A normal subgroup of a group
is a subgroup of
for which the relation "
" of
and
is compatible with the law of composition on
, which in this article is written multiplicatively. The quotient group of
under this relation is often denoted
(said, "
mod
"). (Hence the notation
for the integers mod
.)
An equivalent definition of normal subgroups is this-
is said to be a normal subgroup of a group
if
.Note that this means
but it does not imply that for every
,
.
Description
From the characterizations of relations compatible with left and right translation (see the article on cosets), a subgroup is normal if and only if
is equivalent to
, which is in turn true if and only if
implies
, which is in turn equivalent to its converse (by replacing
,
with
,
).
Note that the relation is compatible with right multiplication for any subgroup
: for any
,
On the other hand, if
is normal, then the relation must be compatible with left multiplication by any
. This is true if and only
implies
Since any element of
can be expressed as
, the statement "
is normal in
" is equivalent to the following statement:
- For all
and
,
,
which is equivalent to both of the following statements:
- For all
,
;
- For all
,
.
By symmetry, the last condition can be rewritten thus:
- For all
,
.
Equivalently, one can say that a normal subgroup is one that is stable under all inner automorphisms.
The intersection of a family of normal subgroups of a group
is a normal subgroup of
. For $xy^{-1} \in {{\rm G}_i$ (Error compiling LaTeX. Unknown error_msg) (for each
) implies
(for each
); hence
implies
.
Examples
In an Abelian group, every subgroup is a normal subgroup. More generally, the center of every group is a normal subgroup of that group.
Every group is a normal subgroup of itself. Similarly, the trivial group is a subgroup of every group.
Consider the smallest nonabelian group, (the symmetric group on three elements); call its generators
and
, with
, the identity. It has two nontrivial subgroups, the one generated by
(isomorphic to
and the one generated by
(isomorphic to
). Of these, the second is normal but the first is not.
If and
are groups, and
is a homomorphism of groups, then the inverse image of the identity of
under
, called the kernel of
and denoted
, is a normal subgroup of
(see the proof of theorem 1 below). In fact, this is a characterization of normal subgroups, for if
is a normal subgroup of
, the kernel of the canonical homomorphism
is
.
Note that if is a normal subgroup of
and
is a normal subgroup of
,
is not necessarily a normal subgroup of
.
Every characteristic subgroup of is a normal subgroup of
.
Group homomorphism theorems
Theorem 1. An equivalence relation on elements of a group
is compatible with the group law on
if and only if it is equivalent to a relation of the form
, for some normal subgroup
of
.
Proof. One direction of the theorem follows from our definition, so we prove the other, namely, that any relation compatible with the group law on
is of the form
, for a normal subgroup
.
To this end, let be the set of elements equivalent to the identity,
, under
. Evidently, if
, then
, so
; the converse holds as well, so
is equivalent to the statement "
". Also, for any
,
so
. Thus
is closed under the group law on
, so
is a subgroup of
. Then by definition,
is a normal subgroup of
.
Theorem 2. Let and
be two groups; let
be a group homomorphism from
to
, and let
be the kernel of
.
- If
is a subgroup of
, then the inverse image
of
under
is a subgroup of
; if
is normal in
, then its inverse image is normal in
. Consequently,
is a normal subgroup of
, and of this inverse image. If
is surjective, then
, and
induces an isomorphism from
to
.
- If
is a subgroup of
, then
is a subgroup of
; if
is normal in
, then
is normal in
. In particular, if
is surjective, then
is normal in
. The inverse image of
under
is
.
Proof. For the first part, suppose are elements of
. Then
, so
is an element of
. Hence
is a subgroup of
. If
is a normal in
, then for all
in
and all
in
,
so
thus
is normal in
. Applying this result to the trivial subgroup of
, we prove that
is normal in
; since the trivial subgroup of
is also a subgroup of
,
is also a normal subgroup of
. If
is surjective, then by definition
. Also, if
and
are elements of
which are congruent mod
, then
, so
. Thus
induces an isomorphism from
to
which is evidently a homomorphism; hence, an isomorphism. This proves the first part of the theorem.
For the second part, suppose that are elements of
. Then
so
is a subgroup of
and of
. Suppose
is normal in
. If
is any element of
, then
so
is normal in
. If
is surjective, then
, so
is normal in
.
Finally, suppose that is an element of
such that
is an element of
. Then for some
,
. Hence
Then
, for some
. Then
. This finishes the proof of the second part of the theorem.
Corollary 3. Let and
be groups; let
be a subgroup of
and
a normal subgroup of
. Let
be a group homomorphism from
to
, with
the kernel of
. Then
is a normal subgroup of
,
is a normal subgroup of
, and
is a normal subgroup of
; furthermore, the quotient groups
,
, and
are isomorphic.
Proof. By theorem 2, is normal in
. Let
be the canonical homomorphism of
into
; let
be the restriction of
to
, and let
. Then
is a surjective homomorphism from
to
, and its kernel is
. Furthermore,
induces a surjective homomorphism from
to $f({\rm G'})/f({{\rm L})$ (Error compiling LaTeX. Unknown error_msg); the kernel of this homomorphism is
. The corollary then follows from theorem 2.
For the following three corollaries, will denote a group, and
a normal subgroup of
, and
the canonical homomorphism from
to
.
Corollary 4. The mapping is a bijection from the set of subgroups of
that contain
to the set of subgroups of
.
Proof. Evidently, if is a subgroup of
containing
, then
is a subgroup of
. If
is a subgroup of
, then
is a subgroup of
containing
, so
is surjective. Finally, since
is the kernel of
,
, so
is injective.
Corollary 5. Let be a subgroup of
containing
. Then
is normal in
if and only if
is normal in
; in this case, the groups
and
are isomorphic.
Proof. Note that . Then by theorem 2, if
is normal in $\lamda(\rm G) = G/N$ (Error compiling LaTeX. Unknown error_msg), then
is normal in
. Conversely, since
is surjective, if
is normal in
, then
is normal in
. Now, suppose that
is normal in
. Let
be the canonical homomorphism of
onto
. Evidently
is a surjective homomorphism from
to
, and the kernel of
is
. Then by theorem 2,
and
are isomorphic.
Corollary 6. Let be a subgroup of
. Then
is a subgroup of
of which
is a normal subgroup, and the groups
and
are isomorphic.
Proof. By theorem 2, is the inverse image of the image of
under
; hence it is a subgroup of
in which
is evidently normal. Let
be the canonical injection of
into
, and let
be the restriction of
to
. Then
is a surjective homomorphism from
to
, and its kernel is
. Hence
and
are isomorphic, as desired.