Amalgamating infinite cyclic groups by multiplication by m and n gives the presentation with relation x^m=y^n

Given: The objects and hypotheses in the example.

Let $G=\langle X\mid R\rangle$ and $H=\langle Y\mid S\rangle$ with disjoint generators, and let $f,h$ embed $K$. If $T$ generates $K$ and words $u_t(X),v_t(Y)$ represent $f(t),h(t)$, then $G\ast_KH\cong\langle X\sqcup Y\mid R\cup S\cup\{u_t v_t^{-1}:t\in T\}\rangle.$ (A free product with amalgamation has the factor presentations plus the amalgamating relations).

The canonical maps $G\to G\ast_KH$ and $H\to G\ast_KH$ are injective. (The factor maps into a free product with amalgamation are injective).

Every class in $W(X)/{\sim}$ contains exactly one reduced word. (Every class in $W(X)/{\sim}$ contains exactly one reduced word).

For every set $X$, the group $F_{\mathrm{word}}(X)=W(X)/{\sim}$ together with $i_{\mathrm{word}}(x)=[x]$ is a free group on $X$ in the sense of def-free-group. (The word-quotient group $W(X)/{\sim}$ satisfies the universal property of the free group on $X$).

Let $F(X)$ be a free group and let $R\subseteq F(X)$ be a set of words, called relations. The group with presentation $\langle X\mid R\rangle:=F(X)/\langle\!\langle R\rangle\!\rangle_{F(X)}$ is the quotient by the normal closure of $R$. The members of $X$ are its generators. In this quotient, every relation in $R$ becomes the identity, as do all consequences forced by normality. (Group presentation by generators and relations).

Let $G$ be a group and $R\subseteq G$. Then $\langle\!\langle R\rangle\!\rangle_G=\left\{g_1r_1^{\varepsilon_1}g_1^{-1}\cdots g_nr_n^{\varepsilon_n}g_n^{-1}:n\in\mathbb N,\ g_i\in G,\ r_i\in R,\ \varepsilon_i\in\{1,-1\}\right\}.$ For $n=0$ the displayed product is the identity. Replacing every conjugator $g_i$ by $g_i^{-1}$ gives the equivalent convention $g_i^{-1}r_i^{\varepsilon_i}g_i$. (The normal closure of $R$ is the set of finite products of conjugates of elements of $R$ and their inverses).