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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 welldefined 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:
$1\longrightarrow G\longrightarrow E\longrightarrow H\longrightarrow 1.$ 
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
(a) $H$ is isomorphic to a subgroup of $E$, thus viewing $H$ as a subgroup of $E$,
(b) $E=GH$, and
(c) $G\cap H=\langle 1\rangle$.
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 5lemma, $e$ is actually an isomorphism. Thus equivalences of extensions are welldefined.

Like split extensions, special extensions are formed when certain conditions are imposed on $G$, $H$, or even $E$:
(a) If all the groups involved are abelian (only that $E$ is abelian is necessary here), then we have an abelian extension.
(b) 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 $E\cong G\times H$.
(c) If $G$ is a cyclic group, then the extensions in question are called cyclic extensions.
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