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 $\bigsqcup_iX_i$. (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 $\widehat u\circ i=u.$ 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 $\langle g \rangle \;=\; \{\, g^{n} \;:\; n \in \mathbb{Z} \,\} ,$ 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. ($\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, 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 $\operatorname{ord}(g) = n$ 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. $\langle g \rangle = \{\, g^{s} : s \in \mathbb{N},\ s < n \,\}$ and $\langle g \rangle \approx n$; so $\langle g \rangle$ is finite with $|\langle g \rangle| = n = \operatorname{ord}(g)$. Infinite order. If $\operatorname{ord}(g) = \infty$ then for $j, k \in \mathbb{Z}$, $g^{j} = g^{k}$ implies $j = k$; so the integer powers of $g$ are pairwise distinct and $\langle g \rangle$ is not finite. (If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ 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).