Finite-order elements of a free product are conjugate into factors

Given: The objects and hypotheses in the statement.

For groups as in def-group, a syllable is a tagged pair $(i,g)$ with $i\in I$ and $g\in G_i\setminus\{e_i\}$. A reduced syllable word is a finite list of syllables, indexed by a natural length as in def-natural-numbers, in which adjacent tags differ. The empty list is allowed. At a concatenation seam, adjacent syllables from the same factor are multiplied and an identity result is deleted; this elementary reduction is repeated until the seam is reduced. (Reduced syllable words in a family of groups).

Every element of $\ast_{i\in I}G_i$ has a unique reduced syllable expression. The identity is represented by the empty word, and no nonempty reduced word represents the identity. (Normal form theorem for free products).

The order of a finite group. Let $G$ be a group (def-group) whose underlying set is finite (def-countable), so that $G \approx n$ for some $n \in \mathbb{N}$ (def-equinumerous). That natural number is unique: if $G \approx n$ and $G \approx n'$ then $n \approx n'$, since $\approx$ is symmetric and transitive, and then $n = n'$ by claim 3 of lem-pigeonhole. The order of $G$ is that unique natural number, written $|G|$. A group is infinite when its underlying set is not finite, and $|G|$ is then not defined. The order of an element. Let $G$ be any group and $g \in G$, with natural powers as in def-group-power. Put $S_g \;:=\; \{\, k \in \mathbb{N} \;:\; k \ge 1 \text{ and } g^{k} = e \,\} \;\subseteq\; \mathbb{N}.$ - If $S_g \ne \varnothing$, the order of $g$ is its least element, $\operatorname{ord}(g) \;:=\; \min S_g \;\in\; \mathbb{N},$ which exists by the well-ordering principle (thm-well-ordering-principle): every nonempty subset of $\mathbb{N}$ has a least element, and that element is unique, being $\le$ every element of $S_g$ and a member of it. We then say $g$ has finite order. - If $S_g = \varnothing$ we say $g$ has infinite order and write $\operatorname{ord}(g) = \infty$, where $\infty$ is a symbol reserved for this case and is not a natural number. No arithmetic is performed with it here. By construction $\operatorname{ord}(g) \ge 1$ whenever it is finite, and $\operatorname{ord}(g) = 1$ exactly when $g = e$, since $g^{1} = g$. Every element of a finite group has finite order. If $G$ is finite then $S_g \ne \varnothing$ for every $g \in G$, by lem-order-of-element-exists, so $\operatorname{ord}(g)$ is a natural number. (The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity).

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).