a repository of mathematical know-how

Revision of New groups from old from Sun, 26/04/2009 - 16:50

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 iconIncomplete 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

  • Free products

  • Amalgamated 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.


A group G is an extension of groups N and Q 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


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


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.


Tensor products are not

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

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

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.