A free product of copies of the infinite cyclic group is a free group

Given: The objects and hypotheses in the statement.

For pairwise disjoint sets $X_i$, the free product of the free groups $F(X_i)$ is a free group on ${⨆}_i{X}_i$. (Free groups on disjoint bases freely multiply to the free group on their union).

A free group on a set $X$ is a group $F(X)$ together with a map $i:X\to F(X)$ such that, for every group $G$ and every function $u:X\to G$, there is a unique group homomorphism $\widehat{u}:F(X)\to G$ satisfying $\widehat{u}\circ i=u\mathrm{.}$ The reduced-word construction supplies such a group; the construction and its universal property are established in thm-reduced-words-form-the-free-group. When no ambiguity arises, $x\in X$ is identified with its image $i(x)$. (Free group on a set of generators).

Let $G$ be a group and $g\in G$, with integer powers as in def-group-power. Then $⟨g⟩=\{g^n:n\in \mathbb{Z}\},$ the cyclic subgroup generated by $g$ (def-generated-subgroup) being exactly the set of integer powers of $g$. Consequently every cyclic group is abelian, and so is every cyclic subgroup of any group. ($⟨g⟩=\{g^n:n\in \mathbb{Z}\}$, and every cyclic group is abelian).

Let $G$ be a group, $g\in G$, and let orders be as in def-order-in-a-group. Throughout, a natural number written where an integer is expected means its image under the embedding $\iota :\mathbb{N}\to \mathbb{Z}$ of lem-nat-embeds-int. Finite order. Suppose $\mathrm{ord}(g)=n$ with $n\in \mathbb{N}$, $n\ge 1$. Then: 1. for every $k\in \mathbb{Z}$, $g^k=e$ if and only if $k=qn$ for some $q\in \mathbb{Z}$, that is, if and only if $n\mid k$ (thm-division-algorithm-in-z); 2. the powers $g^0,g^1,\dots ,g^{n-1}$ are pairwise distinct: if $i,j\in \mathbb{N}$ with $i<n$, $j<n$ and $g^i=g^j$, then $i=j$; 3. $⟨g⟩=\{g^s:s\in \mathbb{N},s<n\}$ and $⟨g⟩\approx n$; so $⟨g⟩$ is finite with $\mathrm{\mid }⟨g⟩\mathrm{\mid }=n=\mathrm{ord}(g)$. Infinite order. If $\mathrm{ord}(g)=\mathrm{\infty }$ then for $j,k\in \mathbb{Z}$, $g^j=g^k$ implies $j=k$; so the integer powers of $g$ are pairwise distinct and $⟨g⟩$ is not finite. (If $\mathrm{ord}(g)=n$ then $g^k=e$ iff $k$ is an integer multiple of $n$, the powers $g^0,\dots ,g^{n-1}$ are distinct, and $⟨g⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^j=g^k$ only for $j=k$).

Let $G$ be a group (def-group) with identity $e$, let $g,h\in G$, and let powers be as in def-group-power. For all $m,n\in \mathbb{Z}$: 1. $g^{m+n}=g^m{g}^n$; 2. $g^{-m}=(g^m{)}^{-1}$; 3. $(g^m{)}^n=g^{mn}$; 4. $g^m{g}^n=g^n{g}^m$: any two powers of one element commute; 5. if $gh=hg$ then $(gh{)}^n=g^n{h}^n$. Claim 5 is false in general without its hypothesis: in a group in which $g$ and $h$ do not commute the equation can fail already at $n=2$, and a witness is recorded on the companion page. Claims 1 and 3 hold in any monoid (def-semigroup-and-monoid) for exponents in $\mathbb{N}$, and so does claim 5 for exponents in $\mathbb{N}$ under the same commuting hypothesis; only the extension to negative exponents needs inverses. (Exponent laws in a group: $g^{m+n}=g^m{g}^n$ and $(g^m{)}^n=g^{mn}$ for all $m,n\in \mathbb{Z}$, and $(gh{)}^n=g^n{h}^n$ when $g$ and $h$ commute).