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 $G\approx n^{\mathrm{\prime }}$ then $n\approx n^{\mathrm{\prime }}$, since $\approx$ is symmetric and transitive, and then $n=n^{\mathrm{\prime }}$ by claim 3 of lem-pigeonhole. The order of $G$ is that unique natural number, written $\mathrm{\mid }G\mathrm{\mid }$. A group is infinite when its underlying set is not finite, and $\mathrm{\mid }G\mathrm{\mid }$ 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 $S_g:=\{k\in \mathbb{N}:k\ge 1\text{ and }{g}^k=e\}\subseteq \mathbb{N}\mathrm{.}$ - If $S_g\mathrm{\ne }\mathrm{\varnothing }$, the order of $g$ is its least element, $\mathrm{ord}(g):=\min S_g\in \mathbb{N},$ 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 $S_g=\mathrm{\varnothing }$ we say $g$ has infinite order and write $\mathrm{ord}(g)=\mathrm{\infty }$, where $\mathrm{\infty }$ is a symbol reserved for this case and is not a natural number. No arithmetic is performed with it here. By construction $\mathrm{ord}(g)\ge 1$ whenever it is finite, and $\mathrm{ord}(g)=1$ exactly when $g=e$, since $g^1=g$. Every element of a finite group has finite order. If $G$ is finite then $S_g\mathrm{\ne }\mathrm{\varnothing }$ for every $g\in G$, by lem-order-of-element-exists, so $\mathrm{ord}(g)$ is a natural number. (The order $\mathrm{\mid }G\mathrm{\mid }$ of a finite group and the order $\mathrm{ord}(g)$ of an element, with $\mathrm{ord}(g)=\mathrm{\infty }$ 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).