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If C is proper in both factors of $G=G_1{\ast }_C{G}_2$, every central element lies in C. In the Bass-Serre tree, a central element fixing a vertex fixes its entire orbit and then the whole tree, hence an edge. A central element acting as a translation would have an invariant axis; minimality forces the tree to be that axis, but a vertex stabilizer then contains a reflection, contradicting centrality of the translation. Thus the center fixes an edge and is contained in its stabilizer C.

// Target: algebraic-topology.bigb

For nontrivial groups $A,B$, the Cayley graph of $A\ast B$ with $S=(A\setminus \{1\})\cup (B\setminus \{1\})$ has the edges of the Bass-Serre tree as its vertices. Two such vertices are adjacent exactly when the tree edges share an endpoint. Mapping each group element to its tree-edge midpoint is an isometry on the vertex metrics and has image within distance $1/2$ of every tree point, hence gives a quasi-isometry of metric graphs. The normal form theorem for a free product identifies $|g{|}_S$ with reduced syllable length. The factor subgroups have diameter one in this metric.

To form the connected sum of knots $K_1\#K_2$, remove a short unknotted arc from each oriented knot in two disjoint balls and join the remaining endpoints by a pair of parallel arcs respecting the orientations. Equivalently, a sphere meeting a knot in two points cuts it into two one-string tangles; close each by an arc in the sphere to recover its summands. Sliding the small joining balls along the knots shows that the isotopy type does not depend on these choices. The unknot is the identity for this operation. A prime knot is a nontrivial knot such that every expression $K=K_1\#K_2$ has an unknot summand.

We first prove additivity of Seifert genus, which also supplies a terminating measure for decomposition. Joining minimal Seifert surfaces along their boundary arcs gives $g_s(K_1\#K_2)\le g_s(K_1)+g_s(K_2)$. For the reverse inequality, let $F$ be a minimal-genus Seifert surface for the sum and $S$ a splitting sphere of a knot. Arrange transverse intersection so that $F\cap S$ consists of one arc joining the two boundary-intersection points and some circles, with the number of circles minimal among genus-minimizing surfaces.

Each circle separates the sphere into two disks; since it is disjoint from the intersection arc, one disk contains neither marked endpoint. Choose an innermost such circle and its disk $D\subset S$, whose interior misses $F$. If its boundary were essential in $F$, compressing $F$ along $D$ would lower the genus of its component with boundary K: for a nonseparating curve genus drops by one, and for a separating essential curve the removed boundaryless side has positive genus. This contradicts minimal genus. Its boundary therefore bounds a disk in $F$. Replace that disk by a small push-off of $D$, removing the intersection circle without changing genus or boundary, again contradicting minimal intersection. Thus no circles remain.

Cutting $F$ along its single intersection arc gives two connected surfaces with boundaries the two summand knots after the closing arcs are added. Euler characteristic shows that their genera add to $g(F)$. Each genus is at least the minimum genus of its summand, soGenus zero means that the knot bounds a disk and is the unknot. Every nontrivial knot therefore has positive genus. Induct on this nonnegative integer: a nontrivial nonprime knot splits into two nontrivial summands, each with strictly smaller genus by the equality above, and induction decomposes both into primes. This proves every knot is a finite connected sum of prime knots, with the unknot the empty sum. Uniqueness is not needed here.

For coprime integers p,q, define the torus knot as the oriented parametrized curveCoprimality makes it an embedded single circle: two parameters with both coordinate phases equal differ by a multiple of $2\pi$. It lies on the standard unknotted torus separating the two solid tori. Negative exponents change the slope/orientation convention; they are allowed. If an exponent is zero, coprimality forces the other to be plus or minus one, giving an unknot.

Assume now $|p|,|q|\ge 2$. We give a group-theoretic proof that torus knots are prime, making the peripheral ingredient explicit. The Seifert-van Kampen theorem applied to the two solid-torus sides of the exterior, joined along the annulus parallel to the knot, givesHere a and b are the two core loops. The element $z=a^p=b^q$ is central. Under the abelianization $G\to \mathbb{Z}$, choose $a\mapsto q$, $b\mapsto p$; these generate the target because p,q are coprime. Thus $z\mapsto pq\ne 0$, so z is nontrivial. A meridian has, up to conjugacy, the form $m=a^u{b}^v$ with $qu+pv=1$. This follows by tracing the meridional disk across the two torus sides, using a dual primitive slope intersecting the knot slope once. Different Bezout solutions change u by a multiple of p and v by the opposite multiple of q, leaving the same word because z is central.

Quotienting by z gives $G/⟨z⟩=C_{|p|}\ast C_{|q|}$. Neither u nor v is zero modulo the corresponding cyclic order, by $qu+pv=1$. Hence the image of m is a cyclically reduced word of two syllables. All its nonzero powers remain reduced, so it has infinite order. In particular G is nonabelian and the knot is not an unknot.

We use the Loop theorem in its usual form: a noninjective boundary fundamental-group map of a three-manifold supplies an embedded compression disk. For a nontrivial knot, its boundary torus is incompressible. To see why, every sphere in its exterior bounds a ball on the side not containing the knot, by the Schoenflies theorem. A compression of the boundary torus would produce a sphere and make the exterior a solid torus. The compression disk has zero homology class in the exterior, so its boundary is the preferred longitude; attaching the annulus from that longitude to the knot gives a spanning disk and an unknot. This proves that nontrivial knot exteriors have incompressible boundary. Their peripheral groups inject as ${\mathbb{Z}}^2$, so their groups strictly contain the cyclic meridian subgroup.

If the torus knot were a connected sum of two nontrivial knots, van Kampen on the decomposing annulus would give $G=G_1{\ast }_{⟨m⟩}{G}_2$, with the meridian subgroup proper in each factor. The center of a proper amalgam lies in the amalgamated subgroup. One proof uses its Bass-Serre tree: a central elliptic element fixes a vertex orbit and its convex hull, which is the whole tree; a central translation would force the minimal tree to be its invariant axis, but a nontrivial vertex reflection reverses that translation. Thus a central element fixes every edge and lies in the edge subgroup.

The central element z would therefore equal $m^k$. Its image in $G/⟨z⟩$ would give ${\overline{m}}^k=1$, forcing $k=0$ because that image has infinite order. This contradicts $z\ne 1$. Consequently

Each factor is a free abelian group of rank two. Take the infinite symmetric generating setThe Bass-Serre tree has an edge $e_g$ joining $gA$ to $gB$ for every $g\in A\ast B$. Indeed, the normal form theorem for a free product gives $A\cap B=\{1\}$, so a nonempty coset intersection consists of exactly one element. Let $m_g$ be the midpoint of $e_g$ and send the Cayley vertex $g$ to $m_g$.

Two distinct edges $e_g,e_h$ share an endpoint precisely when $g^{-1}h\in A\setminus \{1\}$ or $g^{-1}h\in B\setminus \{1\}$. This is exactly adjacency in the Cayley graph for $S$. A path in that graph therefore gives a path between the midpoints of the same length in the tree. Conversely, the unique tree segment between two edge midpoints passes through a succession of adjacent edges and gives a Cayley path of that length. Hence, on vertices,All edge midpoints occur, and every tree point is within $1/2$ of one of them. Extending along Cayley edges gives a quasi-isometry of the metric graphs; ambiguity from folding the cliques at a tree vertex costs only a bounded additive error. This is the free product with factor generating set construction.

For finite generating sets the answer changes. Use $S_0=\{a,a^{\prime },b,b^{\prime }{\}}^{\pm 1}$. The retraction $A\ast B\to A$ killing $B$ sends every generator to a generator of $A$ or to the identity. Therefore the restriction of $d_{S_0}$ to $A$ is exactly the standard ${\ell }^1$ word metric on ${\mathbb{Z}}^2$.

In this grid, take vertices $(0,0),(2N,0),(0,2N)$. Use the two axis segments as two sides, and the path from $(2N,0)$ through $(2N,2N)$ to $(0,2N)$ as the third. All are metric geodesics in the whole Cayley graph, by the retraction argument. The corner $(2N,2N)$ is at distance $2N$ from either of the other sides. These metric geodesic triangles are arbitrarily thick, so this Cayley graph is not a Gromov-hyperbolic metric space.

A tree is $0$-hyperbolic. Hyperbolicity is invariant under quasi-isometry for metric geodesic spaces, and all finite word metrics on this group are equivalent. Consequently no Cayley graph for a finite generating set is quasi-isometric to any tree, including a locally infinite one.

Yes, but there is only one nontrivial isomorphism type that works. If $G=C_2$, thenThe infinite dihedral group contains its infinite cyclic rotation subgroup with index two, so it is amenable by virtually abelian groups are amenable.

Suppose now that $G\ncong C_2$. The action of $G\ast G$ on its Bass-Serre tree is non-elementary: one vertex degree is greater than two, so there are hyperbolic elements with disjoint pairs of endpoints. Suitable powers satisfy the ping-pong lemma on the boundary and generate a copy of $F_2$. Since a subgroup of an amenable group is amenable, an amenable group cannot contain this nonamenable subgroup. The amenability of a free product therefore yieldsfor nontrivial finitely generated $G$.

Direct multiplication gives $A^3=B^2=-I$, so $⟨A⟩$ has order $6$, $⟨B⟩$ has order $4$, and their intersection is $\{I,-I\}$ of order $2$. For the amalgamated free productthe Bass-Serre tree has verticesand edges $S{L}_2(\mathbb{Z})/\{\pm I\}$. Each edge $g\{\pm I\}$ joins $g⟨A⟩$ to $g⟨B⟩$. Thus it is an infinite bipartite tree in which the $⟨A⟩$-vertices have degree $3$ and the $⟨B⟩$-vertices have degree $2$:with a third branch leaving every ${\circ }_3$ vertex and the same pattern continuing on every branch.

Let $g$ be an isometry of the tree $T$. After subdividing edges if necessary, it has no edge inversion. If $g$ fixes a vertex, it is an elliptic isometry of a tree. Otherwise choose a vertex $v$ minimizing the positive integer $d(v,gv)$. The geodesic segments $g^n[v,gv]$ concatenate without backtracking: any backtracking would produce a vertex with smaller displacement. Their union over $n\in \mathbb{Z}$ is therefore a bi-infinite geodesic, the axis of a tree isometry, and $g$ translates it by $d(v,gv)$. This proves the elliptic-hyperbolic classification of an isometry of a tree.

An element of finite order cannot translate a line through a positive distance, since its powers would have unbounded displacement. Every finite-order element of $S{L}_2(\mathbb{Z})$ therefore fixes a vertex of its Bass-Serre tree. Vertex stabilizers are conjugates of $⟨A⟩$ and $⟨B⟩$, so the element is conjugate to a power of $A$ or a power of $B$.

Consider the principal congruence subgroupIt has finite index because $S{L}_2({\mathbb{F}}_3)$ is a finite group. It is torsion-free: by part (b), a finite-order element is conjugate to $A^i$ or $B^j$, while the reductions of $A$ and $B$ modulo $3$ still have orders $6$ and $4$, respectively. Hence no nonidentity power in either vertex group reduces to the identity.

It follows that $\Gamma (3)$ intersects every conjugate of the two vertex stabilizers trivially, so its action on the Bass-Serre tree is free. A group with a free group action on a tree is a free group. Consequently $S{L}_2(\mathbb{Z})$ contains the free subgroup $\Gamma (3)$ of finite index; equivalently, it is a virtually free group.

The infinite dihedral group is $D_{\infty }\cong C_2\ast C_2$, with factors $A=⟨a⟩$ and $B=⟨b⟩$. Its Bass-Serre tree has vertex setand one edge indexed by each $g\in D_{\infty }$, joining $gA$ to $gB$. Since both factors have order two, every vertex has degree two. The connected tree is therefore a bi-infinite line.

The action is cocompact, and its vertex stabilizers are the finite conjugates of $A$ and $B$, so it is proper. By the Milnor–Švarc lemma, an orbit map from $D_{\infty }$ with a word metric to this line is a quasi-isometry. A simplicial bi-infinite line is quasi-isometric to $\mathbb{R}$, hence so is $D_{\infty }$.

Suppose $G=A\ast B$ were a nontrivial free product. Its Bass-Serre tree action has trivial edge stabilizers and no global fixed vertex. Part c makes $a$ elliptic. Since $a$ has infinite order, the fixed set of every nonzero power $a^r$ is a single vertex: it is nonempty, while fixing two vertices would fix the intervening edge and put the infinite-order element $a^r$ in a trivial edge stabilizer.

Let this vertex be $v$. The relation $b{a}^m{b}^{-1}=a^n$ givesand both sides are the singleton $\{v\}$. Thus $b$ also fixes $v$. Since $a$ and $b$ generate the Baumslag-Solitar group, the entire group fixes $v$, contradicting the Bass-Serre action of a nontrivial free product. Hence no such decomposition exists.

The HNN extension is the Baumslag-Solitar groupIts Bass-Serre tree has vertices $G/⟨u⟩$ and oriented edges $G/⟨u⟩$, with the two endpoint maps induced by the identity embedding and the index-two embedding $n\mapsto 2n$. At each vertex there is one incident edge on the identity side and two on the index-two side. The underlying unoriented tree is therefore infinite and $3$-regular, with an orientation in which every vertex has one incoming and two outgoing edges, up to reversing the convention.

WriteIts Bass-Serre tree has verticesand edges $G/C$, where $C=⟨u^2⟩=⟨v^3⟩$; the edge $gC$ joins $g⟨u⟩$ to $g⟨v⟩$. Sincethis is the infinite $(2,3)$-biregular tree: every $u$-type vertex has degree two and every $v$-type vertex has degree three.

Introduce $d=c^2$ and $e=a^2$. The two vertex groupsare Klein bottle groups. In $A$, the subgroup $⟨a^2,d⟩$ is ${\mathbb{Z}}^2$ of index two; in $B$, the subgroup $⟨e,c^2⟩$ is also ${\mathbb{Z}}^2$ of index two. IdentifyinggivesEliminating $d,e$ from this amalgamated free product recovers exactly the two given relators.

The Bass-Serre tree is bipartite with vertex sets $G/A$ and $G/B$ and edge set $G/⟨a^2,c^2⟩$. Both edge-group inclusions have index two, so every vertex has degree two. The tree is therefore a bi-infinite line, with $A$- and $B$-type vertices alternating.