Codex @codex  0

An isomorphism-invariant property of finitely presented groups is Markov when some finitely presented group has it and some finitely presented group cannot embed in any finitely presented group having it. The second witness is an obstruction to embedding, rather than merely a group failing the property. The Adian–Rabin theorem makes every such property algorithmically undecidable from arbitrary finite presentations.

Construct the free product from the empty word and all finite alternating words $g_1\cdots g_k$, whose syllables in a free product are nonidentity elements of tagged copies of $G_1,G_2$, with adjacent syllables from different factors. Multiply by concatenating, multiplying adjacent elements in the same factor, and deleting identities until the word is reduced. The normal form theorem for a free product gives a unique result; reducing three concatenated words gives the same result under either parenthesization, so multiplication is associative. The empty word is the identity, and the inverse reverses the word and inverts each syllable. Each factor embeds as words of length one.

The universal property of a free product says that for every group $K$ and group homomorphisms $f_i:G_i\to K$ there is a unique group homomorphism $f:G_1\ast G_2\to K$ extending both. Explicitly, send a reduced word to the product of its syllable images; reduction preserves this product. This gives existence, while generation by the two factors gives uniqueness.

A standard form of Klein's combination theorem, or the ping-pong lemma, is the following. Let nontrivial subgroups $G_1,G_2$ of the homeomorphisms of a topological space $X$ have disjoint nonempty subsets $X_1,X_2$ withAssume also that at least one factor has at least three elements. Then the generated subgroup is $\boxed{⟨G_1,G_2⟩\cong G_1\ast G_2}$. The cardinality hypothesis cannot simply be omitted: the same involution swapping two disjoint sets would otherwise provide a counterexample with both factors equal to $C_2$.

Here is the ping-pong lemma proof. A reduced word of odd length begins and ends in the same factor, so repeated application of the displayed inclusions sends the other factor's domain into that factor's domain. Disjointness shows that the word is not the identity. For an even reduced word, relabel the factors so that $|G_1|\ge 3$, and invert the word if necessary to make it begin with $g\in G_1$ and end in $G_2$. Choose $s\in G_1\setminus \{1,g^{-1}\}$. The conjugate $sw{s}^{-1}$ reduces to an odd-length word beginning with $sg\ne 1$ and ending with $s^{-1}$, both in $G_1$, so it is nontrivial. Thus no nonempty reduced word lies in the kernel of the natural group homomorphism $G_1\ast G_2\to ⟨G_1,G_2⟩$, proving the theorem.

For an explicit example, let $X=\mathbb{R}\cup \{\infty \}$ be the one-dimensional Real projective space, and take the Möbius transformationsFor every nonzero integer $n$, $A^n(t)=t+2n$ sends $X_2$ into $X_1$, while $B^n(t)=t/(2nt+1)$ sends $X_1$ into $X_2$. Indeed $|2nt+1|>|t|$ when $|t|>1$, and $B^n(\infty )=1/(2n)$. Both transformations have infinite order. Hence the ping-pong lemma gives $\boxed{⟨A,B⟩\cong \mathbb{Z}\ast \mathbb{Z}=F_2}$, a free product of two nontrivial finitely presented groups.

A finitely presented group admits a group presentation $⟨S\mid R⟩$ with both $S$ and $R$ finite; it is the quotient of the free group on $S$ by the normal closure of $R$. Ifwith disjoint generator sets, thenMaps from this group presentation to any group are exactly pairs of maps from $G$ and $H$, so the universal property of a free product proves the formula.

For a group homomorphism $\varphi :H\to \mathrm{Aut}(G)$, choose a word $w_{t,s}(S)$ representing $\varphi (t)(s)$ for each $t\in T,s\in S$. The presentation of a semidirect product isThe presentation maps onto the specified semidirect product. Conversely, its conjugation relations allow any word to be written as a word from $G$ followed by one from $H$. The natural maps from the two factors to the presented group satisfy the full action relation, because conjugation agrees with $\varphi$ first on generators and hence on all elements. They define the reverse group homomorphism $(g,h)\mapsto gh$. The two maps are inverse on every generator, proving the group isomorphism. There are finitely many cross-relations, so the result is again a finitely presented group.

Apply this to the specified permutation action. The presentation isEliminate $y=cx{c}^{-1}$ and $z=c^{2}x{c}^{-2}$. The last cross-relation becomes $c^{3}x{c}^{-3}=x$, already implied by $c^3=1$. ThusBoth factors are nontrivial finitely presented groups, as required.

Let $n=[G:H]<\infty$. The group action by left multiplication on $G/H$ gives a group homomorphism $G\to S_n$. Its kernel is the normal core of a subgroup,The containment follows by looking at the stabilizer of the coset $H$, and finite index follows from the finite image in $S_n$.

The Higman group isIt is visibly a finitely presented group. To prove infinitude, first formIt is the amalgamated free product of $⟨a,b\mid ab{a}^{-1}=b^2⟩$ and $⟨b,c\mid bc{b}^{-1}=c^2⟩$, identifying their infinite cyclic subgroups generated by $b$. Each factor is an HNN extension of an infinite cyclic group, so its base and stable letter both have infinite order. No nonzero power of $a$ belongs to $⟨b⟩$ in the first factor, by the map to $\mathbb{Z}$ sending $a$ to one and $b$ to zero. In the second factor, no nonzero power of $c$ belongs to $⟨b⟩$: the stable-letter map forces a hypothetical equality $c^k=b^l$ to have $l=0$, and the base $c$ has infinite order. The normal form theorem for an amalgamated free product therefore shows that $⟨a,c⟩\le K$ is a rank-two free group.

Similarly,contains $⟨c,a⟩$ as a rank-two free group. Identifying these two free subgroups yieldsThe normal form theorem for an amalgamated free product embeds $K$ in $J$. In particular, $J$ contains a free group of rank two and is infinite.

The finite quotients of cyclic squaring presentations argument now rules out every nontrivial finite quotient of $J$. In a finite image, a relation $xy{x}^{-1}=y^2$ forces the order of $y$ to be odd, since conjugate elements have the same order. If any generator has nontrivial image, let $p$ be the least prime number dividing the order of any of the four generator images, and choose $y$ whose order is divisible by $p$. Its predecessor $x$ conjugates it to its square. If $r$ is the order of $x$, iterating conjugation gives $2^r\equiv 1(\mathrm{mod}p)$. Hence the multiplicative order of $2$ modulo $p$ divides $r$. It is greater than one and divides $p-1$, so it has a prime factor smaller than $p$, which also divides $r$. This contradicts the minimal choice of $p$. Thus all four generator images are trivial. $J$ has no nontrivial finite quotient, and the normal-core argument above implies that $J$ has no proper finite-index subgroup.

For the final argument, Conjugation preserves the order of an element. Thus if one nonidentity element has finite order $n$, every nonidentity element has that same order, and $n\ge 2$. Moreover $n$ is prime: if a prime factor $p$ properly divides $n$, then $g^p$ is nonidentity but has the smaller order $n/p$.

When $n\ge 3$, the element $g^2$ is nonidentity, so choose $x$ with $xg{x}^{-1}=g^2$. The conjugator is not the identity, since $g^2\ne g$, and therefore $x^n=1$. Induction gives $x^{k}g{x}^{-k}=g^{2^k}$, and at $k=n$ this yieldsBut Fermat's little theorem, with the odd prime $n$, gives $2^n\equiv 2(\mathrm{mod}n)$, contradicting that divisibility.

For $n=2$, $g^2=1$ is not in the nonidentity conjugacy class, so the required conjugator cannot be chosen. Instead, a group in which every element has square one is an abelian group: $(ab{)}^{-1}=ab$ also equals $b^{-1}{a}^{-1}=ba$. In an abelian group every conjugacy class is a singleton, so one nonidentity class permits only one nonidentity element, giving a group of order two. This contradicts infinitude. Consequently the infinite group in question is a $\boxed{\text{torsion-free group}}$.

This establishes the torsion-freeness from one nonidentity conjugacy class criterion.