Let $G$ and $H$ be groups. A group $E$ is called an <</SPAN>#110#>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 $E/N$

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. $G\cap H=\langle1\rangle$
  Equivalently, $E$ is a semidirect product of $G$ and $H$ if the short exact sequence $$1\longrightarrow G\longrightarrow E\stackrel{\alpha}{\longrightarrow} H\longrightarrow 1$$ splits. That is, there is a group homomorphism $\phi\colon H\to E$ such that the composition $$H\stackrel{\phi}{\longrightarrow}E\stackrel{\alpha}{\longrightarrow}H$$ 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 $ghg^{-1}=\overline{h}\in H$ 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 $e\colon E_1\to E_2$ such that the following diagram of two short exact sequences is commutative: $$\xymatrix{1\ar@{=}[d]\ar[r]&G\ar@{=}[d]\ar[r]&E_1\ar[d]^e\ar[r]&H\ar@{=}[d] \ar[r]&1\ar@{=}[d]\\1\ar[r]&G\ar[r]&E_2\ar[r]&H\ar[r]&1.}$$ 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.