Free product ($A\ast B$)

The free product $A\ast B$ is the amalgamated free product over the trivial group. Its nonidentity elements have unique reduced alternating normal forms.

Let $G=⟨x,y\mid y^2=1⟩$. For each $n\ge 2$, a free basis of a subgroup of rank $n$ isA covering-space proof uses $r=n-1$ spheres double-covering the real projective plane, with antipodal points $a_j,b_j$ over the wedge point. Add lifted $x$ edges in a cyclic order through the $a_j$, and an $x$ loop at each $b_j$. This is a connected $2r$-sheeted cover of a circle wedged with a projective plane. After collapsing one connecting arc in each simply connected sphere, its fundamental group is that of a graph with $r$ vertices and $2r$ edges, hence free of rank $r+1$.

A spanning tree along the first $r-1$ cyclic edges gives the displayed words: the closing cycle is $x^r$, and the loop reached through $a_j,b_j$ is $x^{j}yxy{x}^{-j}$. Injectivity of the induced fundamental-group map of a covering map proves they form a free basis in $G$.

Suppose $P$ has a soluble word problem for a group and $z$ generates an infinite cyclic group. In $P\ast ⟨z⟩$, for $g\ne 1$ the words $gz$ and $[g,z]$ are cyclically reduced with two and four syllables respectively. Their nonzero powers have syllable lengths $2|k|$ and $4|k|$. Given an input's reduced free product form, its length bounds the possible exponents; compare the finitely many candidates using the base word algorithm. Membership and the exponent in each of these infinite cyclic subgroups are therefore computable, with no algorithm for the order of $g$ required.

A subgroup $A$ is a free factor of $G$ if $G$ is isomorphic to a free product $A\ast B$ by an isomorphism restricting to the inclusion of $A$, for some subgroup $B$. The normal form theorem for a free product ensures that both factors embed. A free factor need not be a normal subgroup.

Given group homomorphisms $f_A:A\to H$ and $f_B:B\to H$, there is exactly one group homomorphism $A\ast B\to H$ extending both. Its value on an alternating normal form is the product of the factor-map values. Thus the free product is the coproduct in the category of groups.

A free product $A\ast B$ is nontrivial as a splitting when both factors differ from the trivial group. Alternating words $(ab{)}^r$ with $a,b\ne 1$ in different factors show that every such splitting gives an infinite group. The distinct normal forms $ab$ and $ba$ also show that the group is not an abelian group.

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.

In a free group, a reduced word has no adjacent inverse letters. Successively cancelling such pairs gives the unique reduced representative in the chosen free generators. In a free product, a reduced word instead uses nonidentity factor elements as syllables in a free product, with consecutive syllables in a free product from different factors; the alternating syllables in a free product are unique.

For an amalgamated free product $A{\ast }_{C}B$, the usual reduced form relative to a common subgroup requires the maps $C\to A$ and $C\to B$ to be injective. Chosen representatives of cosets then distinguish the alternating factors and the remaining common-subgroup element. The exact form depends on the chosen left/right representative convention.

For general group pushouts defined by arbitrary homomorphisms, factors can collapse and this embedded-subgroup normal form is unavailable. For example, take $A=B=C=\mathbb{Z}$, map $C$ identically into $A$, and map it trivially into $B$. The pushout kills all of $A$ and is isomorphic to $B$, so words in nonidentity elements of $A$ cannot retain the usual nontrivial-factor normal form.

Every element of a free product $G\ast H$ has a unique expression $u_1\cdots u_n$ in which each $u_i$ is a nonidentity element of one factor and adjacent syllables belong to different factors. The identity corresponds to the empty word.

A syllable is a nonidentity element of one factor in a free product word. A reduced normal form has successive syllables in different factors; its syllable length is the number of these elements. This length records factor elements, rather than letters in a chosen generating set.