Difference between revisions of "Normal subgroup"
(start of an article; I'll be back to finish it) |
(→Group homomorphism theorems: added some more) |
||
Line 36: | Line 36: | ||
<cmath> xy \equiv ee \equiv e \pmod{\mathcal{R}}, </cmath> | <cmath> xy \equiv ee \equiv e \pmod{\mathcal{R}}, </cmath> | ||
so <math>xy \in {\rm H}</math>. Thus <math>{\rm H}</math> is closed under the group law on <math>{\rm G}</math>, so <math>{\rm H}</math> is a subgroup of <math>{\rm G}</math>. Then by definition, <math>{\rm H}</math> is a normal subgroup of <math>{\rm G}</math>. <math>\blacksquare</math> | so <math>xy \in {\rm H}</math>. Thus <math>{\rm H}</math> is closed under the group law on <math>{\rm G}</math>, so <math>{\rm H}</math> is a subgroup of <math>{\rm G}</math>. Then by definition, <math>{\rm H}</math> is a normal subgroup of <math>{\rm G}</math>. <math>\blacksquare</math> | ||
+ | |||
+ | '''Theorem 2.''' Let <math>{\rm G}</math> and <math>{\rm H}</math> be two groups; let <math>f</math> be a group homomorphism from <math>{\rm G}</math> to <math>{\rm H}</math>, and let <math>{\rm N}</math> be the kernel of <math>f</math>. | ||
+ | * If <math>{\rm H'}</math> is a subgroup of <math>{\rm H}</math>, then the inverse image <math>f^{-1}({\rm H) = G'}</math> of <math>{\rm H'}</math> under <math>{\rm H}</math> is a subgroup of <math>{\rm G}</math>; if <math>{\rm H'}</math> is normal in <math>{\rm H}</math>, then its inverse image is normal in <math>{\rm G}</math>. Consequently, <math>{\rm N}</math> is a normal subgroup of <math>{\rm G}</math>, and of this inverse image. If <math>f</math> is [[surjective]], then <math>f({\rm G'}) = {\rm H'}</math>, and <math>f</math> induces an isomorphism from <math>{\rm G'/N}</math> to <math>{\rm H'}</math>.<br> | ||
+ | * If <math>{\rm G'}</math> is a subgroup of <math>{\rm G}</math>, then <math>f({\rm G'})</math> is a subgroup of <math>{\rm H}</math>; if <math>{\rm G'}</math> is normal in <math>{\rm G}</math>, then <math>f({\rm G'})</math> is normal in <math>f({\rm G})</math>. In particular, if <math>f</math> is surjective, then <math>f({\rm G'})</math> is normal in <math>{\rm H}</math>. The inverse image of <math>f({\rm G'})</math> under <math>f</math> is <math>\rm G'N = NG'</math>. | ||
+ | |||
+ | ''Proof.'' For the first part, suppose <math>a,b</math> are elements of <math>{\rm G'}</math>. Then <math>f(ab) = f(a)f(b) \in {\rm H}</math>, so <math>ab</math> is an element of <math>{\rm G'}</math>. Hence <math>{\rm G'}</math> is a subgroup of <math>{\rm G}</math>. If <math>{\rm H'}</math> is a normal in <math>{\rm H}</math>, then for all <math>a</math> in <math>{\rm G}</math> and all <math>b</math> in <math>{\rm G'}</math>, | ||
+ | <cmath> f(a)f(b)f(a)^{-1} \in {\rm H'}, </cmath> | ||
+ | so | ||
+ | <cmath> aba^{-1} \in f^{-1}(\rm H') = G'; </cmath> | ||
+ | thus <math>{\rm G'}</math> is normal in <math>{\rm G}</math>. Applying this result to the trivial subgroup of <math>{\rm H}</math>, we prove that <math>{\rm N}</math> is normal in <math>{\rm G}</math>; since the trivial subgroup of <math>{\rm H}</math> is also a subgroup of <math>{\rm H'}</math>, <math>{\rm N}</math> is also a normal subgroup of <math>{\rm G'}</math>. If <math>f</math> is surjective, then by definition <math>f(\rm G') = H'</math>. Also, if <math>a</math> and <math>b</math> are elements of <math>{\rm G'}</math> which are congruent mod <math>{\rm N}</math>, then <math>f(ab^{-1}) = f(e)</math>, so <math>f(a) = f(b)</math>. Thus <math>f</math> induces an isomorphism from <math>\rm G'/N</math> to <math>\rm H'</math> which is evidently a homomorphism; hence, an isomorphism. This proves the first part of the theorem. | ||
+ | |||
+ | For the second part, suppose that <math>a,b</math> are elements of <math>{\rm G'}</math>. Then | ||
+ | <cmath> f(a)f(b) = f(ab) \in f({\rm G'}) \subseteq f(\rm G) \subseteq H, </cmath> | ||
+ | so <math>f({\rm G'})</math> is a subgroup of <math>{\rm H}</math> and of <math>f({\rm G})</math>. Suppose <math>{\rm G'}</math> is normal in <math>{\rm G}</math>. If <math>x</math> is any element of <math>{\rm G}</math>, then | ||
+ | <cmath> f(x)f(a)f(x)^{-1} = f(xax^{-1}) \in f(\rm G') , </cmath> | ||
+ | so <math>f( \rm G')</math> is normal in <math>f(\rm G)</math>. If <math>f</math> is surjective, then <math>f(\rm G)= H</math>, so <math>f({\rm G'})</math> is normal in <math>{\rm H}</math>. | ||
+ | |||
+ | Finally, suppose that <math>a</math> is an element of <math>{\rm G}</math> such that <math>f(a)</math> is an element of <math>f({\rm G'}</math>. Then for some <math>b \in \rm G'</math>, <math>f(a) = f(b)</math>. Hence | ||
+ | <cmath> f(ab^{-1}) = f(a)f(b)^{-1} = f(e). </cmath> | ||
+ | Then <math>ab^{-1} = n</math>, for some <math>n\in \rm N</math>. Then <math>a= bn \in \rm G'N = NG'</math>. This finishes the proof of the second part of the theorem. <math>\blacksquare</math> | ||
== See also == | == See also == |
Revision as of 18:08, 8 May 2008
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
.)
Description
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
,
.
Examples
In an Abelian group, every subgroup is a normal subgroup.
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
.
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.