Revision of New groups from old from Sun, 26/04/2009 - 23:27

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.

This article is 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.

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 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.

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 $b^{-1}ab=(a)\phi_b$ .

This section could use additional contributions. We need a good example of a use of a semidirect product. Or a direct product, for that matter!

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.

Extensions

A group $G$ is an extension of a group $Q$ by a group $N$ if there is a short exact sequence

$1\to N\to G\to Q\to 1.$

(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 1

It is a famous and non-trivial fact that there exists a finite group presentation

$Q\cong \langle a_1,\ldots,a_m\mid r_1,\ldots, r_n\rangle$

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 Example 2 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

$1\to N\to F\to Q\to 1.$

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 1 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 $N$ .

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 $F$

Fibred products

The fibred product construction in the category of groups is the same as in the category of sets. If $G_1\to Q$ and $G_2\to Q$ 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,

$G_1{\times_Q} G_2=\{(g_1,g_2)\in G_1\times G_2\mid \phi_1(g_1)=\phi_2(g_2)\}$

Fibred products can be used to improve the finiteness properties of subgroups. Example 1 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 (Greenberg's Theorem states that every finitely generated normal subgroup of a finitely generated free group is of finite index) 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 2

Let $Q$ be a finitely presented group with unsolvable word problem as in Example 1 and let $F\to Q$ 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

$K=(\phi\times\phi)^{-1}(\Delta)$

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

$\{(a_1,a_1),\ldots,(a_m,a_m),(1,r_1),\ldots,(1,r_n)\}$

generates $K$ .

We have proved the following.

Proposition 2 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 $K$ .

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

$B\to A\}$

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

$A\wr B=A^B\rtimes B$

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 3

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.

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.

This section could use additional contributions. It would be appropriate to mention something about the connection between wreath products and the induced representation here.

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 push-out of the corresponding diagram is by definition the amalgamated free product $A*_C B$ . This 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 then the fundamental group of $X$ is a push-out. Therefore, amalgamated free products arise very naturally in topology.

Example 4

Suppose $\Sigma$ is a compact orientable surface and $S^1\to \Sigma$ 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 $S^1\hookrightarrow \Sigma_i$ are injective at the level of $\pi_1$ for $i=0,1$ . (This follows from the classification of surfaces.) Therefore

$\pi_1(\Sigma)\cong\pi_1(\Sigma_0)*_\Z\pi_1(\Sigma_1)$

by the Seifert–van Kampen Theorem.

HNN extensions

Example 4 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 $C\hookrightarrow A$ are both injective homomorphisms. If $A$ has presentation $\langle X\mid R\rangle$ then the Higman–Neumann–Neumann (HNN) extension is

$A*_C\cong\langle X,t \mid R, \{t\phi(c)t^{-1}=\psi(t)\mid c\in C\}\rangle.$

Example 5

Suppose $\Sigma$ is a compact orientable surface and $S^1\to \Sigma$ 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 Möbius band). Then

$\pi_1(\Sigma)\cong\pi_1(\Sigma_0)*_{\langle\gamma\rangle}$

by the Seifert–van Kampen Theorem.

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 Example 5, 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 $A*_C$ .

Comments

Tensor products are not

Sat, 25/04/2009 - 05:55 — emerton

Tensor products are not really defined for groups, but rather for modules
over rings. Abelian groups are $\Z$ -modules, and so tensor products are defined for abelian groups, but this is a construction of a very different flavour
to all the other constructions listed on this page.

Perhaps it would be better to have a comment somewhere on the page to this effect
(i.e. that one can define the tensor product of two abelian groups), and then just
link to the How to use tensor products page for more details.

If there are no objections, I will do this some time soon.

I agree

Sat, 25/04/2009 - 09:28 — JoseBrox

Yes, I was thinking about the tensor product for abelian groups as a special case of a product construction (in Ring theory it is quite usual to think of everything as modules). Feel free to change it as you say, I added it just as a suggestion (I put it on the list because there really isn't any more on the stub at the moment!)

There could be a link

Mon, 11/05/2009 - 05:11 — emerton

There could be a link somewhere among the later examples to Use topology to study your group, although I haven't thought very carefully about where it would sit best.