group extension

Let $G$ and $H$ be groups. A group $E$ is called an extension of $G$ by $H$ if

1. $G$ is isomorphic to a normal subgroup $N$ of $E$, and

2. $H$ is isomorphic to the quotient group $E/N$.

The definition is well-defined and it is convenient sometimes to regard $G$ as a normal subgroup of $E$. The definition can be alternatively defined: $E$ is an extension of $G$ by $H$ if there is a short exact sequence of groups:

$$1⟶G⟶E⟶H⟶1.$$

In fact, some authors define an extension (of a group by a group) to be a short exact sequence of groups described above. Also, many authors prefer the reverse terminology, calling the group $E$ an extension of $H$ by $G$.

Remarks

• Given any groups $G$ and $H$, an extension of $G$ by $H$ exists: take the direct product of $G$ and $H$.

• An intermediate concept between an extension a direct product is that of a semidirect product of two groups: If $G$ and $H$ are groups, and $E$ is an extension of $G$ by $H$ (identifying $G$ with a normal subgroup of $E$), then $E$ is called a semidirect product of $G$ by $H$ if

  1. $H$ is isomorphic to a subgroup of $E$, thus viewing $H$ as a subgroup of $E$,

  2. $E=GH$, and

  3. $G\cap H=⟨1⟩$.

  Equivalently, $E$ is a semidirect product of $G$ and $H$ if the short exact sequence

  $$1⟶G⟶E\stackrel{\alpha }{⟶}H⟶1$$

  splits. That is, there is a group homomorphism $\varphi :H\to E$ such that the composition

  $$H\stackrel{\varphi }{⟶}E\stackrel{\alpha }{⟶}H$$

  gives the identity map. Thus, a semidirect product is also known as a split extension. That a semidirect product $E$ of $G$ by $H$ is also an extension of $G$ by $H$ can be seen via the isomorphism $h\mapsto hG$.

  Furthermore, if $H$ happens to be normal in $E$, then $E$ is isomorphic to the direct product of $G$ and $H$. (We need to show that $(g,h)\mapsto gh$ is an isomorphism. It is not hard to see that the map is a bijection. The trick is to show that it is a homomorphism, which boils down to showing that every element of $G$ commutes with every element of $H$. To show the last step, suppose $gh{g}^{-1}=\overline{h}\in H$. Then $gh=\overline{h}g$, so $gh{\overline{h}}^{-1}=\overline{h}g{\overline{h}}^{-1}=\overline{g}\in G$, or that $h{\overline{h}}^{-1}=g^{-1}\overline{g}$. Therefore, $h=\overline{h}$.)

• The extension problem in group theory is the classification of all extension groups of a given group $G$ by a given group $H$. Specifically, it is a problem of finding all “inequivalent” extensions of $G$ by $H$. Two extensions $E_1$ and $E_2$ of $G$ by $H$ are equivalent if there is a homomorphism $e:E_1\to E_2$ such that the following diagram of two short exact sequences is commutative:

  According to the 5-lemma, $e$ is actually an isomorphism. Thus equivalences of extensions are well-defined.

• Like split extensions, special extensions are formed when certain conditions are imposed on $G$, $H$, or even $E$:

  1. If all the groups involved are abelian (only that $E$ is abelian is necessary here), then we have an abelian extension.

  2. If $G$, considered as a normal subgroup of $E$, actually lies within the center of $E$, then $E$ is called a central extension. A central extension that is also a semidirect product is a direct product. Indeed, if $E$ is both a central extension and a semidirect product of $G$ by $H$, we observe that $(g\overline{h})h{(g\overline{h})}^{-1}=\overline{h}h{\overline{h}}^{-1}\in H$ so that $H$ is normal in $E$. Applying this result to the previous discussion and we have $E\cong G\times H$.

  3. If $G$ is a cyclic group, then the extensions in question are called cyclic extensions.

Title	group extension
Canonical name	GroupExtension
Date of creation	2013-03-22 15:24:25
Last modified on	2013-03-22 15:24:25
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	11
Author	CWoo (3771)
Entry type	Definition
Classification	msc 20J05
Related topic	HNNExtension
Defines	split extension
Defines	abelian extension
Defines	central extension
Defines	cyclic extension
Defines	extension problem