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[QUICK DESCRIPTION]

This page lists various constructions of new examples of groups and subgroups from known examples, and gives some brief notes of what each construction is good for.

[note article incomplete]The idea is to give an example of its use for each construction.  I intend to finish this off eventually, but if anyone cares to step in in the meantime, feel free!  Also, each construction probably needs its own article.[/note]

[GENERAL DISCUSSION]

This is a list of various standard constructions of groups and subgroups.

* Direct products
* Semidirect products
* Extensions
** Central extensions
* Fibre products
* Wreath products
* Amalgamated free products
** Free products
* HNN extensions
* Graphs of groups

[remark tensor] Tensor products are another form of product, defined in general not for groups but for modules.  As abelian groups are $\Z$-modules, this construction can be seen as a way of constructing abelian groups.[/remark]

===Direct products===

Given groups $A$ and $B$, the simplest way we might think to construct a product group is by component-wise multiplication on the set $A \times B$. This does give rise to a well-defined group operation, and the resulting group is called the direct product of $A$ and $B$. We can naturally identify $A$ with $A \times \{1_B\} \subset A \times B$ and $B$ with $\{1_A\} \times B \subset A \times B$, so we generally just consider $A$ and $B$ themselves to be subgroups of $A \times B$.

[note section contributions wanted]Need to add in some examples. We should also give the internal form of a direct product -- which I think should be motivated by "how do we know when a group is the direct product of two of its subgroups?" Come to think of it, that would make an interesting page by itself...[/note]

===Semidirect products===

The group law on the direct product $A\times B$ is determined by the requirement that $b^{-1}ab=a$ for every $a\in A$ and $b\in B$.  That is, $B$ acts on $A$ by conjugation, and the action is trivial.   The semidirect product construction derives from the observation that any action of $B$ on $A$ could be used to construct a new group in the same way.  Let $\phi$ be a right-action of $B$ on $A$, so for any $b\in B$ we have an automorphism $\phi_b$ of $A$ that acts on the right, sending an element $a$ to $(a)\phi_b$.  The ''semidirect product'' $A\rtimes_\phi B$ is in bijection with the set of pairs $ab$ where $a\in A$ and $b\in B$, and the group law is determined by the requirement that $b^{-1}ab=(a)\phi_b$.

Often the action $\phi$ is suppressed.  By construction, the subgroup $A$ of $A\rtimes B$ is normal, and the quotient is isomorphic to $B$.  So $A\rtimes B$ is an extension of $B$ by $A$.  In fact, any ''split'' extension of $B$ by $A$ is a semidirect product.

[EXAMPLE semidirect]
Two geometric examples of semidirect products are the [[w:dihedral group]] $D_n$, which is $Z_n\rtimes Z_2$ and the [[w:Euclidean group]], which is $\mathbb{R}^n\rtimes O(n)$.  For the dihedral group, the normal subgroup can be interpreted as the group of rotations by $2\pi/n$ and the $Z_2$ factor is some reflection.  The action $\phi_b$ of reflections on the rotations simply reverses the direction of rotation.  To see this, note that rotating by $2\pi/n$ then reflecting is the same as reflecting and then rotating by $-2\pi/n$.

For the Euclidean group, the normal subgroup is $n$-dimensional translations, and the action $\phi_b$ is the same as rotating the coordinates of the translation.
[/EXAMPLE]

[note section contributions wanted]These are the first examples which came to mind.  I didn't discuss them in much detail, so feel free to expand / correct.  We probably want an example of a direct product first.[/note]

===Extensions===

A group $G$ is an ''extension'' of a group $Q$ by a group $N$ if there is a short exact sequence
[maths]
1\to N\to G\to Q\to 1.
[/maths]
(In other words, the $Q$ is the quotient of $G$ by $N$.)  The following example takes advantage of two features of extensions.

* The subgroup $N$ is to $G$ as the trivial subgroup is to $Q$.
* Often $G$ can be chosen to have better properties than $Q$.  In particular, any presentation for $Q$ corresponds to an extension where $G$ is free.

[EXAMPLE extension]
It is a famous and non-trivial fact that there exists a finite group presentation
[math]
Q\cong \langle a_1,\ldots,a_m\mid r_1,\ldots, r_n\rangle
[/math]
in which the word problem is unsolvable.  That is, there is no algorithm that will tell you whether or not a given word in the generators represents the identity in $Q$.  (We will not use the fact that $n$ is finite in this example.  But it will be important in [ref Example #fibre] below.)

We will use group extensions to produce a different pathology in a much better behaved group---a free group.

Let $F$ be the free group on $\{a_1,\ldots,a_m\}$.   The relations $r_1,\ldots,r_n$ can be thought of as elements of $F$.   By the universal property of free groups, the obvious map $\{a_1,\ldots, a_m\}\to Q$ extends to a surjection $F\to Q$.  The kernel of this surjection is precisely $N=\langle\langle r_1,\ldots, r_m\rangle\rangle$, the normal subgroup of $F$ generated by the relations.  That is, we have a short exact sequence
[math]
1\to N\to F\to Q\to 1.
[/math]
The fact that the word problem is unsolvable in $Q$ can now be restated precisely as the assertion that there are normal subgroups of $F$ with unsolvable membership problem.

[proposition]
There is a normal subgroup $N$ of $F$ with the property that there is no algorithm to determine whether or not a given element of $F$ is in $N$.
[/proposition]

So we see that the existence of the highly pathological group $Q$ corresponds to a different sort of pathological behaviour in the well-behaved group $F$
[/EXAMPLE]

===Fibred products===

The fibred product construction in the category of groups is the same as in the category of sets.  If $\phi_1:G_1\to Q$ and $\phi_2:G_2\to Q$ are surjections then the ''fibred product'' of $\phi_1$ and $\phi_2$ is the subgroup of $G_1\times G_2$ defined as the preimage of the diagonal subgroup of $Q\times Q$ under the map $\phi_1\times \phi_2$.  That is,
[math]
G_1{\times_Q} G_2=\{(g_1,g_2)\in G_1\times G_2\mid \phi_1(g_1)=\phi_2(g_2)\}
[/math]

Fibred products can be used to improve the finiteness properties of subgroups.  [ref Example #extension] showed how to construct a non-trivial subgroup $N$ of a free group with unsolvable membership problem.  Although $N$ was finitely generated as a normal subgroup of $F$, it is a consequence of Greenberg's Theorem [add]{{(Greenberg's Theorem states that every finitely generated normal subgroup of a finitely generated free group is of finite index)}}[/add] that $N$ is not finitely generated as a group.  In fact, every finitely generated subgroup of a free group has solvable membership problem.

In the following example, we will use a fibred product to construct a finitely generated subgroup of a direct product of two free groups that has unsolvable membership problem.

[EXAMPLE fibre]
Let $Q$ be a finitely presented group with unsolvable word problem as in [ref Example #extension] and let $\phi:F\to Q$ be the quotient map derived from the presentation.  Let $K$ be the fibre product of two copies of $\phi$, a subgroup of $F\times F$.  Then
[math]
K=(\phi\times\phi)^{-1}(\Delta)
[/math]
where $\Delta$ is the diagonal subgroup of $Q\times Q$.  The membership problem for $\Delta$ in $Q\times Q$ is unsolvable, (indeed, $(g,1)$ is an element of $\Delta$ if and only if $g$ is trivial in $Q$) and this translates precisely to the statement that the membership problem for $K$ in $F\times F$ is unsolvable.  But the finite set
[math]
\{(a_1,a_1),\ldots,(a_m,a_m),(1,r_1),\ldots,(1,r_n)\}
[/math]
generates $K$.

We have proved the following.

[proposition]
There is a finitely generated subgroup $K$ of $F\times F$ with the property that there is no algorithm to determine whether or not a given element of $F\times F$ is in $K$.
[/proposition]
[/EXAMPLE]

===Wreath products===

A wreath product is a special case of a semidirect product.  We will first restrict our attention to the wreath product of two finite groups $A$ and $B$.  The set of set maps
[math]
A^B=\{f:B\to A\}
[/math]
is a group (multiplication comes from multiplication in $A$) and is naturally equipped with a right-action of $B$, namely the action by left translation.  (It is easy to get confused by the fact that left translation is a right action!) We can think of $A^B$ as the direct sum of copies of $A$, indexed by the elements of $B$, and $B$ acts by permuting the factors.  This is precisely the data needed for a semidirect product construction.

The wreath product of $A$ by $B$ is defined to be
[math]
A\wr B=A^B\rtimes B
[/math]
where $B$ acts on $A^B$ by left translation.

When $A$ or $B$ may be infinite, we define $A^B$ to be precisely those set maps that equal the identity on all but finitely many elements of $B$.  This has the effect that $A^B$ is isomorphic to the direct ''sum'', rather than the direct product, of $|B|$ copies of $A$. Our first example of a wreath product shows how to construct a 2-generator group with an abelian subgroup of infinite rank.

[EXAMPLE wreath1]
The group $\Z^\Z$ is the direct sum of countably many copies of $\Z$, and so can be thought of as the group of biinfinite sequences of integers that are equal to zero in all but finitely many coordinates.  It admits an action of $\Z$, where the integer $n$ acts by moving the $i$th coordinate to the $i+n$th coordinate.  The orbit of the sequence that is $0$ in every non-zero coordinate and $1$ in the $0$th coordinate generates $\Z^\Z$.  (Here the fact that $\Z^\Z$ is the direct sum, rather than the direct product, is important.)  The resulting semidirect product is precisely the wreath product $\Z\wr\Z$.  It contains the infinite-rank abelian group $\Z^\Z$ as a subgroup, and is generated by just two elements. 
[/EXAMPLE]

More generally, given any transitive action of $B$ on a set $X$, one can define the wreath product $A\wr (B,X)$ to be the semidirect product of $A^X$ and $B$, where as before $B$ acts on $A^X$ by left-translation.

[note section contributions wanted]It would be appropriate to mention something about the connection between wreath products and the induced representation here.[/note]

===Amalgamated free products===

Amalgamated free products are push-outs in the category of groups.  If $C\hookrightarrow A$ and $C\hookrightarrow B$ are both injective then the  amalgamated free product $A*_C B$ is defined by the property that the diagram

[image afp.gif|centre|]

is a push out.  That is,  $A*_C B$ is the freest possible group that contains $A$ and $B$ as subgroups in which the two copies of $C$ are identified.

The Seifert--van Kampen Theorem asserts that if a path-connected topological space $X$ can be decomposed as the union of two closed, path-connected subsets whose intersection is also path-connected then the fundamental group of $X$ is a push-out.  Therefore, amalgamated free products arise very naturally in topology.

[EXAMPLE separating]
Suppose $\Sigma$ is a compact orientable surface and $\gamma:S^1\to \Sigma$ is a simple closed curve that is not homotopic to a point.  Suppose further that $\gamma$ is separating---that is, $\Sigma\smallsetminus\mathrm{im}\gamma$ has two path-components; we shall denote their closures by  $\Sigma_0$ and $\Sigma_1$.  Because $\gamma$ is not homotopic to a point, the natural inclusions $\gamma_i:S^1\hookrightarrow \Sigma_i$ are injective at the level of $\pi_1$ for $i=0,1$.  (This follows from the classification of surfaces.)  Therefore
[math]
\pi_1(\Sigma)\cong\pi_1(\Sigma_0)*_\Z\pi_1(\Sigma_1)
[/math]
by the Seifert--van Kampen Theorem. 
[/EXAMPLE]

One of the most important results for proving theorems about amalgamated free products is the Normal Form Theorem, which gives a criterion that determines when elements of $A*_C B$, expressed as products of elements of the images of $A$ and $B$, are nontrivial.  It is a consequence that the maps $A\to A*_C B$ and $B\to A*_C B$ are both injective.  For more details, see the article [[How to prove facts about graphs of groups]].  We will describe the normal form theorem in the important special case of free products below.

====Free products====

If $C$ is the trivial group then the amalgamated product of $A$ and $B$ over $C$ is called the ''free product'' of $A$ and $B$, denoted $A*B$.  This is the freest possible group that contains both $A$ and $B$.  The Normal Form Theorem for free products determines when elements of $A*B$ are non-trivial.  Note that it is immediate from the definition that $A*B$ is generated by the union of $A$ and $B$.

[theorem freenormal]Consider $g\in A*B$ of the form  
[math]
g=a_0 b_1 a_1 b_2 \ldots b_n a_n.
[/math]
If $g=1$ then either some $a_i=1$ for $0<i<n$ or some $b_i=1$.
[/theorem]

===HNN extensions===

[ref Example #separating] explains what happen at the level of $\pi_1$ when you cut a surface $\Sigma$ along a separating curve.  But the separating hypothesis is rather unnatural---it makes just as much sense to cut $\Sigma$ along a non-separating curve $\gamma$.  What happens in this case?  The answer is that $\pi_1(\Sigma)$ decomposes as an HNN extension.

Suppose $\phi,\psi:C\hookrightarrow A$ are both injective homomorphisms.  If $A$ has presentation $\langle X\mid R\rangle$ then the ''Higman--Neumann--Neumann (HNN) extension'' is 
[math]
A*_C\cong\langle X,t  \mid R, \{t\phi(c)t^{-1}=\psi(t)\mid c\in C\}\rangle.
[/math]
The generator $t$ is called the ''stable letter''

[EXAMPLE nonseparating]
Suppose $\Sigma$ is a compact orientable surface and $\gamma:S^1\to \Sigma$ is a simple closed curve that is not homotopic to a point.  Suppose further that $\gamma$ is non-separating, so $\Sigma\smallsetminus\mathrm{im}\gamma$ has one path-component $\Sigma_0$, and two-sided (that is, $\gamma$ is not the core of a M\"obius band).  Then
[math]
\pi_1(\Sigma)\cong\pi_1(\Sigma_0)*_{\langle\gamma\rangle}
[/math]
by the Seifert--van Kampen Theorem. 
[/EXAMPLE]

There are two things to notice about this definition.  The first is that, ''a priori'', it seems to depend on the chosen presentation for $A$---however, one can show that the definition is in fact independent of this choice.  More importantly, $A*_C$ is rather poor notation as it does not specify the maps $\phi$ and $\psi$.  Often, as in [ref Example #nonseparating], the two maps are implicit.  One way of getting round this ambiguity is to set $\chi=\phi^{-1}\circ\psi$ and write $A*_\chi$ instead of $A*_C$.  Alternatively, we sometimes write $A*_{\phi(C)\sim\psi(C)}$.

The analogue of the Normal Form Theorem for amalgamated products in the context of HNN extensions is [[w:HNN extension|Britton's Lemma]], which gives a criterion for when elements of an HNN extension are trivial.  In particular, Britton's Lemma implies that the natural homomorphism $A\to A*_C$ is injective.

As is apparent from the definition, HNN extensions force elements to be conjugate.  In a similar spirit to [ref Example #wreath1], one can use an HNN extension to reduce the number of generators required.  However, HNN extensions are more flexible than semidirect products as they only require two isomorphic subgroups, rather than an automorphism of the whole group.  The following example is related to the original application of HNN extensions by Higman, Neumann and Neumann.

[EXAMPLE embedding]
We will give a proof of the following.

[theorem embeddingthm]
Every countable group embeds into a 3-generator group.
[/theorem]

Let $G=\{1=g_0,g_1,g_2,\ldots\}$ be any countable group.  In order to be able to apply an HNN extension, we first need to modify $G$ so that it has two isomorphic subgroups.  Consider $G'=G*\langle s\rangle$, the free product of $G$ with an infinite cyclic group.  For each $n\in\N$, let $s_n=g_ns^n$ and let $S_n=\{s_m\mid m\geq n\}$.  It follows from the normal form theorem for free products that the subgroup generated by $S_n$ is isomorphic to the free group on $S_n$, for each $n>0$.  In particular, $\langle S_1\rangle\cong\langle S_2\rangle$.  So we can construct the HNN extension
[math]
G''=G'*_{\langle S_1\rangle\sim\langle S_2\rangle}.
[/math]
If $t$ is the stable letter of the HNN extension then we have
[math]
g_{n+1}=t g_ns^n t^{-1}s^{-n-1}
[/math]
for each $n$, and so $G' '$ is generated by $\{g_1,s,t\}$ as required.
[/EXAMPLE]

===Graphs of groups===

In [ref Example #separating] and [ref Example #nonseparating], we saw what happens at the level of the fundamental group when an orientable surface is cut along an embedded curve.  But what if we want to cut along several disjoint curves at once?  (The union of finitely many disjoint curves is called a ''multicurve''.)

[EXAMPLE multicurve]
Let $\Sigma$ be a compact, orientable surface and let $\gamma_1,\ldots,\gamma_n:S^1\to\Sigma$ be a finite set of disjoint simple closed curves.  Assume that no $\gamma_i$ is homotopic to a point.  The complement of the images of the $\gamma_i$ is a disjoint union of connected components $\Sigma_j$:

[math]
\Sigma\smallsetminus\mathrm{im}\gamma_1\cup\ldots\cup\mathrm{im}\gamma_n=\Sigma_1\cup\ldots\cup\Sigma_m.
[/math]

Correspondingly, the fundamental group of $\pi_1(\Sigma)$ decomposes as the fundamental group of a [[w:graph of groups]].  The ''vertex groups'' are the fundamental groups of the $\Sigma_j$, the ''edge groups'' are the cyclic subgroups generated by the $\gamma_i$, and the ''underlying graph'' has one vertex for each $\Sigma_j$ and one edge for each $\gamma_j$, with the obvious incidence relations.
[/EXAMPLE]

The definition of a graph of groups is rather technical, so we will not give it here.  But the main ideas are captured by [ref Example #multicurve].  Note that the multicurve might have happened to consist of just one connected component. Likewise, the definition of a graph of groups generalizes both the definitions of amalgamated products and HNN extensions.  These correspond to the two cases in which the underlying graph has precisely one edge: the underlying graph of an amalgamated product is two vertices joined by an edge; the underlying graph of an HNN extension is a loop consisting of one vertex and one edge.

In [ref Example #multicurve], rather than cutting $\Sigma$ along the multicurve $\gamma_1\cup\ldots\cup\gamma_n$ we could have cut $\Sigma$ along $\gamma_1$, then cut the remaining pieces along $\gamma_2$, and so on.  This operation expresses $\pi_1(\Sigma)$ iteratively as a sequence of amalgamated products and HNN extensions.  Likewise, any graph of groups can be thought of as built up from its vertex groups by taking iterative amalgamated free products and HNN extensions over its edge groups.  So any group that you construct as a graph of groups could also have been constructed using amalgamated products and HNN extensions.

Graphs of groups really come into their own if you want to understand subgroups of amalgamated products and HNN extensions.  To see why, let's return to the example of a curve on a surface.

[EXAMPLE multicurve2]
Let $\Sigma$ be a compact, orientable surface and let $\gamma$ be an embedded simple closed curve that is not homotopic to a point.  Let
[math]
\rho:\widehat{\Sigma}\to\Sigma
[/math]
be a finite-sheeted covering map.  The pre-image $\rho^{-1}(\gamma)$ is easily seen to be a multicurve
[math]
\hat{\gamma}=\gamma_1\cup\ldots\cup\gamma_n
[/math]
and by the homotopy lifting property no $\gamma_i$ is homotopic to a point.

Now let us translate this into group theory.  The curve $\gamma$ decomposes $\pi_1\Sigma$ as either an amalgamated free product or an HNN extension (depending on whether $\gamma$ is separating or non-separating).  But the  multicurve $\hat{\gamma}$ decomposes $\pi_1\widehat{\Sigma}$ as a graph of groups.  Of course we could focus on just one of the $\gamma_i$, which decomposes $\pi_1\widehat{\Sigma}$ as an amalgamated product or HNN extension, but that would involve making an unnatural choice.  Furthermore, the multicurve $\hat{\gamma}$ has the attractive property that the restriction of the covering map $\rho$ to a connected component of $\widehat{\Sigma}\smallsetminus\hat{\gamma}$ is a covering map onto a connected component of $\Sigma\smallsetminus\gamma$.
[/EXAMPLE]

[ref Example #multicurve2] suggests that a subgroup of an amalgamated product or HNN extension naturally decomposes as a graph of groups.  To be precise, the following is true for amalgamated products, and a similar theorem holds for HNN extensions (or indeed any graph of groups).

[theorem]
If $G=A*_C B$ and $H$ is a subgroup of $G$ then $H$ decomposes as a graph of groups, such that every vertex group is conjugate to a subgroup of $A$ or $B$ and every edge group is conjugate to a subgroup of $C$.
[/theorem]

The most usual modern technique for proving theorems about graphs of groups (including amalgamated products and HNN extensions) makes use of Serre's observation that every graph of groups corresponds to an action of its fundamental group on a tree.  This tree is called the Bass--Serre tree, and is the object of study in [[To prove a theorem about graphs of groups translate the problem into a question about group actions on trees|Bass--Serre Theory]].

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