**Theorem:** A prime can be written as the sum of two squares if and only if .

There exist many different proofs of this theorem, but a surprising one is exposed by D. Zagier in his article *A one-sentence proof that every prime is a sum of two squares*. Our note is dedicated to this proof.

First of all, it is easy to prove that the sum of two squares is either congruent to , or modulo . Thus, because is necessarily odd, if it can be written as the sum of two squares then .

Conversely, let be a prime satisfying , and define the set

.

Notice that is nonempty since : there exists a such that

.

Now, let be the map defined by

It is not difficult to show that is well-defined on , because for all . Furthermore, an easy calculation proves that the image of is included into .

Now, we want to prove that is an involution with exactly one fixed point.

**The map is an involution.** Let . If , then implies

;

if , then implies

;

if , then implies

.

Therefore, ie. is an involution.

**The map has only one fixed point.** Clearly, if then is a fixed point of . Conversely, let be a fixed point of . If then is a solution of

,

hence , which is impossible. If then is a solution of

,

hence . Thus, implies . Because is prime, necessarily , and we deduce that .

Finally, if then is a solution of

,

hence , which is impossible. Therefore, has exactly one fixed point: .

In order to conclude the proof of our theorem, we need the following lemma:

**Lemma:** Let be a set and an involution. Then .

**Proof.** Let be a maximal subset of satisfying . Then

,

which implies . This proves the lemma.

Now we are able to conclude the proof of our theorem. Let be a new map defined by . Clearly, is a well-defined involution of . By applying our previous lemma twice, we deduce that

.

In particular, is nonempty. Let be one of its points. Then

.

This proves that can be written as the sum of two squares, and the proof is complete.

]]>

- two generic orientation-preserving homeomorphisms of the circle generate a non abelian free subgroup,
- a generic continuous function is nowhere differentiable.

See the previous notes Free groups acting on the circle and Almost all continuous functions are nowhere differentiable. Now, we want to prove:

**Theorem:** If , the space is simply connected.

Fix a base point and an enumeration . Given two closed curves satisfying , we want to prove that and are homotopic (relatively to ) in . Let

.

It is a closed subset of the Banach space of the continuous functions with respect to the supremum norm , so that , endowed with the distance induced by , is a complete metric space. Now, let

.

Because , the space is simply connected, so that . In fact, it is not difficult to convince ourself that is a dense open subset of . A possible argument is the following:

First, if and if denotes the distance between and (which is positive since is compact), then the ball in of radius and centered at is clearly included into . Therefore, is open in .

Then, let be a bump function whose support is included into and satisfying . Now, given a homotopy , define the perturbation

.

Of course, and . Furthermore, if is small enough and well chosen, then . This proves that is dense in .

Now, it is sufficient to apply Baire category theorem to conclude that the intersection is non empty. But any element of this intersection defines a homotopy between and satisfying , that is to say a homotopy in .

Therefore, we have proved that any two closed curves in based at are homotopic in . This shows that the group is trivial, ie., the space is simply connected.

]]>and let denote the set of the “true” dominoes of size 2, that is

Is it possible to tile using only ? Surprisingly, this problem can be solved thanks to the van Kampen diagrams used in group theory.

Let be a group presentation. A *van Kampen diagram *, relative to , is a finite connected planar graph whose edges are oriented and labelled by the generators , such that the word labelling the boundary of any cell of (ie., any bounded component of ) is a relator . For convenience, we call the word labelling the boundary of the unbounded component of the *boundary word* of .

For example, if we consider the presentation , the following graph defines a van Kampen diagram (relative to ) whose boundary word is .

**Theorem 1:** Let be a group with a presentation . A given word written over is equal to in if and only there exists a van Kampen diagram , relative to , whose boundary word is .

**Proof.** Suppose that there exists a van Kampen diagram whose boundary word is . We easily prove that in by induction on the number of cells of . If has just one cell, then its boundary word is a relator of by definition, so that that trivially. If has at least two cells, then we can write where corresponds to the intersection between the boundary path of and the boundary of one of its cells . Let be the relator associated to . Removing to , we obtain a new van Kampen diagram whose boundary word is . By our induction hypothesis, we conclude that

in our group .

Conversely, suppose that in . Thus, as an element of the free group , the word belongs to the normal subgroup generated by , so we can write

for some and . Now consider the following van Kampen diagram:

Clearly, it defines a van Kampen diagram whose boundary word is .

Now, we associate a group presentation to any set of dominoes . Let be two fixed letters. To any dominoe , we define a word written over following the rule illustrated by

Let denote the group presentation and the underlying group. For example,

with the notation . Now, we can notice that a tiling of some chessboard using produces a van Kampen diagram relative to . For example, the tiling

of using produces the van Kampen diagram

Thus, if we defined the *boundary word *of a chessboard as the word obtained by following the same rule as the one used for the dominoes, we deduce from Theorem 1:

**Theorem 2:** If a chessboard can be tiled by a set of dominoes , then its boundary word is equal to in the group .

As a concrete application, we are now able to solve our initial problem:

**Theorem 3:** can be tiled by if and only if and don’t have the same parity.

**Proof.** As we already noticed,

.

The boundary path of is

.

Notice that and belongs to the center of . Thus, if and are both odd,

,

and if they are both even,

.

Now we want to prove that and in . For this purpose, let us consider the morphism

onto the infinite dihedral group . It is clear that and

in . Therefore, these words are not equal to in as well. According to Theorem 2, we conclude that cannot be tiled by if and have the same parity.

Conversely, suppose without loss of generality that is even and odd. Now, the particular tiling

clearly yields a tiling of by .

In fact, Theorem 3 can be deduced from the following tricky argument: First, it is obvious that if can be tiled by then must have an odd number of squares. Thus, if and are both odd, cannot be tiled by . Secondly, if we color the squares of in black and white exactly as a true chessboard and if are both even, then is obtained from a chessboard by removing two squares of the same color, say two black squares. But, if we color the dominoes of so that they contain a square of each color, we deduce that, if could be tiled by , it would contain the same number of black and white squares. Therefore, we conclude that cannot be tiled by if and are both even.

In the same way, if we denote by the chessboard

then we can prove:

**Theorem 4:** can be tiled by if and only if or is odd.

However, the argument using van Kampen diagrams can be adapted successfully to other tiling problems, where our previous trick does not work any more. Let denote the following set of dominoes:

Then, the associated group presentation is

.

Following the proof of Theorem 3, we get:

**Theorem 5:** can be tiled by if and only if .

The proof is left as an exercice.

This point of view was introduced by Conway and Lagarias, in their article *Tiling with polyominoes and combinatorial group theory*, and was also exploited by Thurston in *Conway’s tiling groups*. See also Hitchman’s article, *The topology of tile invariants*, and references therein for more information.

]]>

In particular, we will prove that the groups for , for , and for are not 3-manifold groups.

**Definitions:** A *handlebody of genus * is an orientable 3-manifold (with boundary) obtained by attaching “handles” to a 3-ball. Roughly speaking, it is a full closed surface of genus .

A way to construct 3-manifolds (without boundary) is to glue two handlebodies and of some genus along their boundaries by a homeomorphism . We say that the triple is a *Heegaard splitting of genus * of the resulting manifold.

**Theorem 1:** Every closed 3-manifold admits a Heegaard splitting.

**Sketch of proof.** Let be a closed 3-manifold, which will be thought of as a three-dimensional simplicial complex. Let denote a small neighborhood of the 1-skeleton of and its complement. If is a 3-simplex, then and are illustrated by the following picture:

Notice that, if is the graph whose vertices are the centers of the 3-simplices of and whose edges link two vertices corresponding to adjacent 3-simplices, then may be thought of as a neighborhood of .

For , let denote a maximal subtree of . Then, may be thought of as a neighborhood of , which is homeomorphic to a 3-ball, with one handle for each edge of the complement . Therefore, is a handlebody of genus , since, if and denotes respectively the number of vertices and edges of a graph,

and

imply that

.

Therefore, we have proved that is the union of two handlebodies which intersect along their boundaries. To conclude, we need to argue that and have the same genus, or equivalently, that .

Let denote respectively the number of vertices, edges, faces and 3-cells of . By construction,

and

.

However, by Poincaré’s duality, , hence

.

The proof is complete.

From a Heegaard splitting of genus , it is possible to find a presentation of the fundamental group of the associated 3-manifold :

We apply van Kampen’s theorem to the decomposition . The handlebodies and have a free group of rank as fundamental groups, and they meet in along a closed surface of genus . Let be a generating set of . Then,

,

where (resp. ) is a generating set of (resp. ).

It is worth noting that such a presentation has the same number of generators and relations; we say that the presentation is *balanced*. Therefore, we have proved:

**Corollary 2:** Every 3-manifold group admits a balanced presentation.

So, a natural question is: which groups admit a balanced presentation? In fact, this question is linked to *(co)homology of groups*, which we define now.

A space is *aspherical* whenever for all ; according to Whitehead’s theorem, it amounts to require the universal covering to be contractible. An interesting property of such spaces is that they are uniquely determined, up to homotopy, by their fundamental groups, that is to say, two aspherical spaces are homotopy equivalent if and only if their fundamental groups are isomorphic. Given a group , a * space* or a *classifying space* for is an aspherical space whose fundamental group is isomorphic to ; according to our previous remark, such a space is uniquely determined by , up to homotopy. In particular, this allows us to define the *-th (co)homology group* of as the -th (co)homology group of such a classifying space.

In order to justify that (co)homology groups are always defined, we need to prove that every group has a classifying space. First, to a given presentation of we associate a two-dimensional CW-complex as follow: Define as a bouquet of circles, each labelled by a generator of . Then, for every relation , glue a -cell along the path labelling by . Using van Kampen’s theorem, it is not difficult to notice that the fundamental group of our complex is isomorphic to . However, it may not be aspherical. Now, let us define an increasing sequence of CW-complexes such that and for all (such a sequence will define a *Postnikov tower*). The spaces are defined inductively in the following way: Let be a generating set for ; then, define by gluing -cells to using the . By construction, , and we claim that the embedding is -injective for all by cellular approximation. Finally, let . By construction, is aspherical, and because have been constructed just by gluing -cells to with , we deduce that (since the fundamental group depends only on the -skeleton of a CW-complex). Therefore, is a classifying space of .

For more information on Postnikov towers, see section 4.3 of Hatcher’s book, Algebraic Topology.

Now, we are ready to linked the *deficiency* of a presentation, that is the number of its generators minus the number of its relations, with the first and second homology groups of the underlying group. The result below is mentioned by Brown in the fifth exercice of chapter II.5 of his book *Cohomology of groups*. Here is the topological approach he suggests:

**Theorem 3:** Let be a presentation of a group with and . Let also denote . Then .

**Proof.** Let be the CW-complex associated to the given presentation. According to what we have said above, we may add -cells to , with , in order to “kill” higher homotopy groups, ie., our new CW-complex is aspherical. Moreover, because we did not modify the -squeleton of , is isomorphic to . Therefore, is a , hence .

Notice that , hence

.

On the other hand, we clearly have an epimorphism , therefore:

**Corollary 4:** If a group admits a balanced presentation, then .

Thus, now we have a necessary criterion to determine whether or not a group admits a balanced presentation. The Proposition below applies this criterion to the right-angled Artin groups (we defined them in our previous note Some SQ-universal groups).

**Proposition 5:** A right-angled Artin group admits a balanced presentation if and only if has more vertices than edges.

**Sketch of proof.** To the canonical presentation of is associated a cube complex , called *Salvetti complex*: has only one point, one edge for each generator of , and generators span an -cube if and only if they pairwise commute. Now, it can be proved that the Salvetti complex is nonpositively curved so that its universal cover is CAT(0); because a CAT(0) space is contractible, we deduce that is a classifying space of . Moreover, it is not difficult to prove that the -th homology group of is free on the set of -cubes. Therefore, corresponds to the number of vertices of , and to its number of edges.

Thus, if admits a balanced presentation, according to Theorem 3, must have more vertices than edges. The converse is obvious.

For more information on CAT(0) geometry, see Bridson and Haefliger’s book, *Metric spaces of nonpositive curvature*.

Now, we are ready to prove that the groups mentionned at the beginning of this note do not admit a balanced presentation, and thus are not the fundamental groups of closed 3-manifolds.

**Corollary 6:** Let . Then admits a balanced presentation if and only if .

**Proof.** is the right-angled Artin group associated to the complete graph . Because has vertices and edges, it is sufficient to apply Proposition 5 to conclude.

**Corollary 7:** Let . Then admits a balanced presentation if and only if and or and .

**Proof.** is the right-angled Artin group associated to the bipartite complete graph . Because has vertices and edges, it is sufficient to apply Proposition 5 to conclude.

**Corollary 8:** Let . Then admits a balanced presentation if and only if and or and .

**Proof.** Let denote the graph with vertices and no edges. Then is the right-angled Artin group associated to the join . Because has vertices and edges (here, we suppose ), it is sufficient to apply Proposition 5 to conclude.

**Theorem:** .

As noticed in the previous note Some SQ-universal groups, as a corollary we get that is SQ-universal, that is every countable group is embeddable into a quotient of .

Usually, the theorem above is proved thanks to a natural action of on the hyperbolic plane by Möbius transformations. However, as noticed by Roger Alperin in his article published in *The American Mathematical Monthly*, it is sufficient to make act on the set of irrational numbers via:

.

Notice that for all precisely because is irrational. The argument below may be thought of as an application of ping-pong lemma (that we already met in the note Free groups acting on the circle in order to find free subgroups).

First, it is a classical exercice to prove that is generated by the two following matrices:

and .

For convenience, let . Then and also generate , and consequently their images in generate it; furthermore, (resp. ) has order two (resp. order three) in . Notice that

, and .

Now, let and denote the set of positive and negative irrationals respectively. Clearly,

and .

To conclude, we want to prove that, for any alternating word from and , in .

**Case 1:** has odd length. Then either begins and ends with a , hence , or begins and ends with a , hence . In particular, we deduce that in .

**Case 2:** has even length. Without loss of generality, we may suppose that begins with a ; otherwise, just conjugate by . Then either begins with a and ends with a , hence

,

or begins with a and ends with a , hence

.

In either case, we deduce that , so in . The proof is complete.

]]>**Property 1:** A SQ-universal group contain a non-abelian free group (of countable rank).

**Proof.** Let be a SQ-universal group. In particular, there exists a quotient containing a free group of infinite rank . Let be a lift of in . If there existed a non-trivial relation between the ‘s then the same relation would hold between the ‘s in . Therefore, is a free subgroup of infinite rank in .

**Property 2:** A countable SQ-universal group contains uncountably many normal subgroups.

**Proof.** Let be a countable SQ-universal group. Every quotient of contains only countably many two-generated subgroups. However, we know that there exist uncountably many two-generated groups up to isomorphism: see for instance the notes Amalgamated products and HNN extensions (I): A theorem of B.H. Neumann or Cantor-Bendixson rank in group theory: A theorem of B.H. Neumann. Therefore, must have uncountably many quotients, and a fortiori uncountably many normal subgroups.

Although SQ-universality seems to be a very strong property, many groups turn out to be SQ-universal. We already saw in previous notes that the free group of rank two is SQ-universal; in fact, we gave two proofs, the one in Amalgamated products and HNN extensions (I): A theorem of B.H. Neumann, and the other, completely elementary, in A free group contains a free group of any rank. Other examples come from the following trivial observation: a group with a SQ-universal quotient is itself SQ-universal. Therefore, any group with a non-abelian free quotient is SQ-universal; in particular, we deduce that any non-abelian free group, of any rank, is SQ-universal. Together with the following result, we will be able to exhib a lot of other examples of SQ-universal groups.

**Theorem 3:** A group containing a SQ-universal subgroup of finite-index is itself SQ-universal.

For an (almost) elementary proof, see P. Neumann’s article, *The SQ-universality of some finitely presented groups*, J. Aust. Math. Soc. 16, 1 (1973) .

**Example 1:** Let be a closed surface. Then either is SQ-universal or is a sphere , a projective plane , a torus or a Klein bottle .

Notice that we already dealt with surface groups in the previous note On subgroups of surface groups. In particular, we know that , , are not SQ-universal. We also know that contains as a subgroup of finite-index two, so it cannot be SQ-universel.

From now on, let be a closed surface different from the surfaces mentioned above. If is non-orientable, there exist a finite-cover where is orientable; in particular, is a subgroup of finite-index in , so, according to Theorem 3, we may suppose that is orientable, that is

for some . Noticing that the following quotient is free,

,

we deduce that is SQ-universal.

**Example 2:** A right-angled Artin groups is either free abelian or SQ-universal.

The right-angled Artin group associated to a finite simplicial graph is defined by the presentation

,

where and denote respectively the set of vertices and edges of . In particular, if is a complete graph with vertices, then is the free abelian group ; if is a graph with vertices and no edges, then is the free group .

For another example, if is the graph given above, then

is isomorphic to .

We claim that, if is not a complete graph, then has a non-abelian free quotient, and in particular is a SQ-universal group.

Because is not complete, there exist two vertices not linked by an edge. Let denote the subgraph induced by and its neighbors, and the subgraph induced by . Then and are non-empty, since and , they cover , and is not connected, because and are separated by . Now, quotienting by the subgroup

,

we get the free group . For an explicit example, let us consider the following graph:

So, the associated right-angled Artin group is

.

Quotienting by the subgroup , we get the free group .

**Example 3:** If and are two finite groups with and , then the free product is a SQ-universal group.

We proved in the previous note A free group contains a free group of any rank that

is a free basis of the derived subgroup of . Furthermore, the quotient is finite, so that is a non-abelian free subgroup of finite index in . From Theorem 3, we deduce that is SQ-universal.

For example, is SQ-universal, because it has the SQ-universal group

as a quotient. (See An elementary application of ping-pong lemma for an elementary proof of this fact.)

]]>**Theorem: **Let be a group. If the center is a finite-index subgroup of , then the commutator subgroup is finite.

We begin with two easy lemmae. The first one is left to the reader as an exercice: it is just a computation. In the sequel, we use the notation .

**Lemma 1:** For all , .

**Lemma 2:** If is a subgroup of finite index , then .

**Proof.** Notice that, because is a subgroup of finite index , for all we have . Therefore,

.

Then,

.

Using Lemma 1, we conclude that

.

**Proof of Theorem:** Let us suppose that is a subgroup of finite index .

It is not difficult to notice that in whenever and in . Therefore, has at most commutators.

Now, any can be written as a product of commutators

.

Suppose that is as small as possible, and assume by contradiction that . Therefore, because there exist at most commutators, the product contains some commutator, say , at least times. We claim that it is possible to write

for some new commutators . For example, if , notice that

.

Thanks to Lemma 1, we know that each is again a commutator. We just proved our claim for . The general case follows by induction.

From and Lemma 2, we deduce that can be written as a product of commutators, a contradiction with the minimality of .

Therefore, we just proved that any element of can be written as a product of at most commutators. Because there exist at most commutators, we deduce that the cardinality of is bounded above by ; in particular, is finite.

**Corollary 1:** If has only finitely many commutators, then is finite.

**Proof.** Let denote the commutators of and be the subgroup generated by the ‘s. Notice that , so it is sufficient to prove that is finite to conclude. Furthermore, according to our previous theorem, it is sufficient to prove that is a finite-index subgroup of .

If denotes the centralizer of in , notice that

.

Therefore, it is sufficient to prove that each is a finite-index subgroup of . But is finite if and only if has only finitely-many conjugacy classes. Because for all , and because has only finitely-many commutators, it is clear that has finitely-many conjugacy classes, concluding our proof.

**Corollary 2:** The only infinite group all of whose non-trivial subgroups are finite-index is .

**Proof.** Let be such a group. Let and be a set of representatives of the cosets of . For each , is a finite-index subgroup of by assumption; but it is also a subgroup of the centralizer , so is a finite-index subgroup of . Therefore, because generates , the center

is a finite-index subgroup of . According to our previous theorem, is finite, and by assumption, any subgroup of is either trivial, or finite-index and in particular infinite since so is. Therefore, is trivial, that is is abelian.

From the classification of finitely-generated abelian groups, it is not difficult to prove that the only abelian group all of whose non-trivial subgroups are finite-index is .

Notice however that there exist non-cyclic groups all of whose normal subgroups have finite-index: they are called *just-infinite groups*. The infinite dihedral group is such an example.

**Corollary 3:** A torsion-free virtually infinite cyclic group is infinite cyclic.

**Proof.** Let be a torsion-free group with an infinite cyclic subgroup of finite index . As above, if denotes a set of representatives of the cosets of then generates . Noticing that , we deduce that is of finite index in , so the centralizer has a finite index in . Again, we deduce that the center

is a finite-index subgroup of . According to our previous theorem, is finite, and in fact trivial because is torsion-free, so is abelian. From the classification of finitely-generated abelian groups, it is not difficult to prove that the only abelian torsion-free virtually cyclic group is .

In fact, using Stallings theorem, it is possible to prove more generally that any torsion-free virtually free group turns out to be free.

A group is abelian if and only if if and only if . Therefore, a way to say that is *almost* abelian is to state that is *big* (eg. is a finite-index subgroup) or that is *small* (eg. is finite). Essentially, our main theorem proves that the first idea implies the second. Although the converse does not hold in general, it holds for finitely-generated groups:

**Theorem:** Let be a finitely-generated group. If is finite then is a finite-index subgroup.

**Proof.** Let be a finite generating set of . Then

.

To conclude, it is sufficient to prove that each is a finite-index subgroup, that is equivalent to say that each has finitely-many conjugacy classes. Noticing that for all , , we deduce the latter assertion because is finite.

]]>**Definition:** Let be a geodesic space and . For convenience, let be the only (up to isometry) simply connected Riemannian surface of constant curvature and let denote its diameter; therefore, is just the hyperbolic plane , the euclidean plane or the sphere with a rescaling metric and is finite only if .

For any geodesic triangle – that is a union of three geodesics , and – of diameter at most , there exists a *comparison triangle* in (unique up to isometry) such that

, and .

We say that satisfies * inequality* if for every

.

We say that is a space if every geodesic triangle of diameter at most satisfies inequality, and that is *of curvature at most * if it is locally .

This definition of curvature bounded above is good enough to agree with sectional curvature of Riemannian manifolds: A Riemannian manifold is of curvature at most if and only if its sectional curvature is bounded above by . For more information, see Bridson and Haefliger’s book, *Metric spaces of nonpositive curvature*, theorem I.1.A6.

From now on, let us consider a *nice* kind of action on geodesic spaces:

**Definition:** A group acts *geometrically* on a metric space whenever the action is properly discontinuous and cocompact.

Such actions are fundamental in geometric group theory, notably because of Milnor-Svarc theorem: if a group acts geometrically on a metric space , then is finitely-generated and there exists a quasi-isometry between and . See [BH, proposition I.8.19].

Let us say that a group is if it acts geometrically on some space. Notice that, if is a space, then is a space. Therefore, a group is either or or .

According to the remark made at the beginning of this note, and properties should give more information on our group than property. In fact, we are able to prove the following result:

**Theorem:** Any finitely-presented group is .

**Sketch of proof.** Let be a finitely-presented group. If is a Cayley complex of , it is know that the natural action is geometric – see Lyndon and Schupp’s book, *Combinatorial group theory*, section III.4. Now, the barycentric subdivision of is a flag complex of dimension two. According to Berestovskii’s theorem mentionned in [BH, theorem II.5.18], the right-angled spherical complex associated to is a (complete) space. Of course, the action is again geometric, since the underlying CW complex is just .

Another important way to link a group with a geometric space is to write as a fundamental group. As above, we can prove the following result:

**Theorem:** Any group is the fundamental group of a space . Moreover, if is finitely-presented, can be supposed compact.

**Sketch of proof.** Let be the CW complex associated to a presentation of ; then and is compact if the presentation were finite – see Lyndon and Schupp’s book, *Combinatorial group theory*, section III.4. As above, the right-angled spherical complex associated to the barycentric subdivision of is a space, and its fundamental group is again isomorphic to , since the underlying CW complex is just .

On the other hand, many properties are known for groups; the usual reference on the subjet is Bridson and Haefliger’s book, *Metric spaces of nonpositive curvature*. Furthermore, is not difficult to prove that groups are Gromov-hyperbolic. However, it is an open question to know whether or not hyperbolic groups are , or even .

Let us conclude this note by noticing that, for a fixed space , it is possible that only few groups are able to act on it. For example, we proved in our previous note Brouwer’s Topological Degree (IV): Jordan Curve Theorem that, for any even number , and are the only groups acting freely by homeomorphisms on the -dimensional sphere .

]]>

.

Of course, such a cancellation property does not hold in general, for example

whereas and are not isomorphic. However, cancellation property turns out to hold for some classes of groups. In his article On cancellation in groups, Hirshon proves that finite groups are *cancellation groups*, that is to say, for every groups , if with finite, then . Below, we present an elementary proof of a weaker statement due to Vipul Naik:

**Theorem:** Let be three finite groups. If then .

**Proof.** For any finite groups and , let denote the number of homomorphisms from to and denote the number of monomorphisms from to . Notice that

Let be three finite groups such that . Then

for any finite group , since . Using , it is easy to deduce that for any finite group by induction on the cardinality of . Hence

Therefore, there exists a monomorphism from to . From , we deduce that and have the same cardinality, so we conclude that and are in fact isomorphic.

Because a classification of finitely-generated abelian groups or of divisible groups is known, it is easy to verify that cancellation property holds also for such groups (the classification of divisible groups was mentionned in a previous note). However, in his book *Infinite abelian groups*, Kaplansky mentions that the problem is still open for the class of all abelian groups.

We conclude our note by an example where the cancellation property does not hold even if the groups are all finitely-presented.

Let be the following presentation

.

Clearly, we have the following decomposition:

.

Now, we set and . Because commutes with and , we deduce that . Then,

and

.

Now, let us notice that

,

hence . Therefore,

.

Using , it is not difficult to notice that the relation can be written as . Then, we deduce that

so that the relation can be removed from the presentation. Finally,

.

Therefore, we find another decomposition of as a direct product:

.

So . To conclude, it is sufficient to prove that and are not isomorphic. Notice that they are both a semi-direct product .

Let be a morphism. It is not difficult to show that a torsion element of has to be conjugated to , so, without loss of generality, we may suppose that for some . Then, for some . Let be the canonical projection sending to and to . If is onto, then so is : because and , we deduce that . On the other hand,

,

hence or . Therefore, has to divide and we deduce that . We conclude that cannot be an isomorphism.

Finally, we proved that

,

where and , but and are not isomorphic.

In particular, we deduce that:

**Corollary:** is not a cancellation group.

]]>

**Theorem:** The additive groups , , and are all isomorphic.

Indeed, they are all -vector spaces of dimension . Another less known example is:

**Theorem:** and are isomorphic.

**Proof.** Let be a basis of as a -vector space; without loss of generality, suppose . We can write as a disjoint union where and . Let and denote the subspaces generated by and respectively; as additive groups, they are both isomorphic to since they are -vector spaces of dimension . Finally, let denote the subspace generated by . Then

.

Now, notice that the morphism

gives an isomorphism , and the morphism

gives an isomorphism . Therefore, we deduce that

.

Noticing that the polar decomposition

gives an isomorphism , we conclude the proof.

Nevertheless, the converse of our criterium does not hold, that is there exist two non-isomorphic vector spaces such that the associated groups are isomorphic: and give such an example as -vector spaces. In particular, it is worth noticing that the choice of the field is crucial.

**Nota Bene:** In this note, we used the axiom of choice dramatically, assuming that every vector space has a basis. In fact, the axiom of choice turns out to be necessary at least to prove the isomorphism , as noticed by C.J. Ash in his article *A consequence of the axiom of choice*, J. Austral. Math. Soc. 19 (series A) 1975 (pp. 306-308). More precisely, the author shows that the isomorphism implies the existence of a set of reals not Lebesgue measurable, concluding thanks to a model for ZF due to Solovay in which every subset of reals is Lebesgue measurable.