Difference between revisions of "Group extension"
(started page) |
(added some) |
||
Line 1: | Line 1: | ||
Let <math>F</math> and <math>G</math> be [[group]]s. An '''extension''' of <math>G</math> by <math>F</math> is a solution to the problem of finding a group <math>E</math> that contains a [[normal subgroup]] <math>F'</math> [[isomorphic]] to <math>F</math> such that the [[quotient group]] <math>E/F'</math> is isomorphic to <math>G</math>. | Let <math>F</math> and <math>G</math> be [[group]]s. An '''extension''' of <math>G</math> by <math>F</math> is a solution to the problem of finding a group <math>E</math> that contains a [[normal subgroup]] <math>F'</math> [[isomorphic]] to <math>F</math> such that the [[quotient group]] <math>E/F'</math> is isomorphic to <math>G</math>. | ||
− | More specifically, an extension <math>\mathcal{E}</math> of <math> | + | More specifically, an extension <math>\mathcal{E}</math> of <math>G</math> by <math>F</math> is a triple <math>(E,i,p)</math> where <math>E</math> is a group, <math>i</math> is an [[injective]] group [[homomorphism]] of <math>F</math> into <math>E</math>, and <math>p</math> is a [[surjective]] homomorphism of <math>E</math> onto <math>G</math> such that the [[kernel]] of <math>p</math> is the image of <math>i</math>. Often, the extension <math>\mathcal{E}</math> is written as the diagram <math>\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G</math>. |
An extension is '''central''' if <math>i(F)</math> lies in the [[center]] of <math>E</math>; this is only possible if <math>F</math> is [[abelian group | commutative]]. It is called ''trivial'' if <math>E = F \times G</math> (the direct product of <math>F</math> and <math>G</math>), <math>i</math> is the canonical mapping of <math>F</math> into <math>F\times G</math>, and <math>p</math> is the [[projection homomorphism]] onto <math>G</math>. | An extension is '''central''' if <math>i(F)</math> lies in the [[center]] of <math>E</math>; this is only possible if <math>F</math> is [[abelian group | commutative]]. It is called ''trivial'' if <math>E = F \times G</math> (the direct product of <math>F</math> and <math>G</math>), <math>i</math> is the canonical mapping of <math>F</math> into <math>F\times G</math>, and <math>p</math> is the [[projection homomorphism]] onto <math>G</math>. | ||
Line 9: | Line 9: | ||
A ''retraction'' of an extension is a homomorphism <math>r: E\to F</math> such that <math>r\circ i</math> is the identity function on <math>F</math>. Similarly, a ''section'' of an extension is a homomorphism <math>s : G \to E</math> such that <math>p \circ s = \text{Id}_G</math>. | A ''retraction'' of an extension is a homomorphism <math>r: E\to F</math> such that <math>r\circ i</math> is the identity function on <math>F</math>. Similarly, a ''section'' of an extension is a homomorphism <math>s : G \to E</math> such that <math>p \circ s = \text{Id}_G</math>. | ||
− | {{ | + | == Equivalence of Extensions == |
+ | |||
+ | '''Theorem 1.''' Let <math>\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G</math> and <math>\mathcal{E}' : F \stackrel{i'}{\to} E' \stackrel{p'}{\to} G</math> be extensions of <math>G</math> by <math>F</math>. Let <math>u : E \to E'</math> be a morphism of extensions. Then <math>u</math> is an isomorphism of <math>E</math> onto <math>E'</math>. In other words, every extension morphism is an extension isomorphism. | ||
+ | |||
+ | ''Proof.'' Suppose <math>a,b</math> are elements of <math>E</math> such that <math>u(a)=u(b)</math>. Then | ||
+ | <cmath> p(a) = (p' \circ u)(a) = (p' \circ u)(b) = p(b) , </cmath> | ||
+ | so <math>ab^{-1} \in \text{Ker}(p) = \text{Im}(i)</math>. Let <math>g \in F</math> be an element such that <math>i(g) = ab^{-1}</math>. Then | ||
+ | <cmath> e_{E'} = u(a)u(b)^{-1} = u(ab^{-1}) = (u \circ i)(g) = i'(g) . </cmath> | ||
+ | It follows that <math>g</math> is the identity of <math>F</math>, so <math>ab^{-1}</math> is the identity of <math>F</math> and <math>a=b</math>. | ||
+ | |||
+ | Let <math>a'</math> be in <math>E'</math>; since <math>p</math> is surjective, there exists <math>a\in F</math> such that <math>p'(a') = p(a) = p'(u(a))</math>. Then | ||
+ | <cmath> a' u(a)^{-1} \in \text{Ker}(p') = \text{Im}(i') = \text{Im}(u \circ i) \subseteq \text{Im}(u). </cmath> | ||
+ | Thus there exists <math>b \in F</math> such that <math>u(b) = a'u(a)^{-1}</math>; then <math>u(ba) = a'</math>. Thus <math>u</math> is surjective. <math>\blacksquare</math> | ||
+ | |||
+ | '''Theorem 2.''' Let <math>\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G</math> be an extension of <math>G</math> by <math>F</math>. Then the following are equivalent: | ||
+ | # <math>\mathcal{E}</math> is the trivial extension; | ||
+ | # <math>\mathcal{E}</math> admits a retraction; | ||
+ | # <math>\mathcal{E}</math> admits a section <math>s</math> such that <math>s(G)</math> lies in the [[centralizer]] of <math>i(F)</math>. | ||
+ | |||
+ | ''Proof.'' If <math>\mathcal{E}</math> is the trivial extension, then the projection onto <math>F</math> and the canonical injection of <math>G</math> into <math>E</math> show that conditions 2 and 3 are satisfied. If <math>\mathcal{E}</math> has a retraction <math>r</math>, then the mapping <math>(r,p): E \to F \times G</math> is an extension morphism, so <math>\mathcal{E}</math> is isomorphic to the trivial extension. If (3) holds, then the mapping <math>(f,g) \mapsto i(f)s(g)</math> is an extension morphism <math>F\times G \to \mathcal{E}</math>, so again <math>\mathcal{E}</math> is isomorphic to the trivial extension of <math>G</math> by <math>F</math>. <math>\blacksquare</math> | ||
+ | |||
+ | Note that an extension <math>F \stackrel{i}{\to} E \stackrel{p}{\to} G</math> may be nontrivial, but <math>E</math> may still be isomorphic to <math>F \times G</math>. | ||
+ | |||
+ | == See also == | ||
+ | |||
+ | * [[Semi-direct product]] | ||
+ | * [[Simple group]] | ||
+ | * [[Normal subgroup]] | ||
+ | * [[Quotient group]] | ||
[[Category:Group theory]] | [[Category:Group theory]] |
Latest revision as of 19:53, 27 May 2008
Let and
be groups. An extension of
by
is a solution to the problem of finding a group
that contains a normal subgroup
isomorphic to
such that the quotient group
is isomorphic to
.
More specifically, an extension of
by
is a triple
where
is a group,
is an injective group homomorphism of
into
, and
is a surjective homomorphism of
onto
such that the kernel of
is the image of
. Often, the extension
is written as the diagram
.
An extension is central if lies in the center of
; this is only possible if
is commutative. It is called trivial if
(the direct product of
and
),
is the canonical mapping of
into
, and
is the projection homomorphism onto
.
Let and
be two extensions of
by
. A morphism of extensions from
to
is a homomorphism
such that
and
.
A retraction of an extension is a homomorphism such that
is the identity function on
. Similarly, a section of an extension is a homomorphism
such that
.
Equivalence of Extensions
Theorem 1. Let and
be extensions of
by
. Let
be a morphism of extensions. Then
is an isomorphism of
onto
. In other words, every extension morphism is an extension isomorphism.
Proof. Suppose are elements of
such that
. Then
so
. Let
be an element such that
. Then
It follows that
is the identity of
, so
is the identity of
and
.
Let be in
; since
is surjective, there exists
such that
. Then
Thus there exists
such that
; then
. Thus
is surjective.
Theorem 2. Let be an extension of
by
. Then the following are equivalent:
is the trivial extension;
admits a retraction;
admits a section
such that
lies in the centralizer of
.
Proof. If is the trivial extension, then the projection onto
and the canonical injection of
into
show that conditions 2 and 3 are satisfied. If
has a retraction
, then the mapping
is an extension morphism, so
is isomorphic to the trivial extension. If (3) holds, then the mapping
is an extension morphism
, so again
is isomorphic to the trivial extension of
by
.
Note that an extension may be nontrivial, but
may still be isomorphic to
.