Amalgamating C_2 inside C_4 and C_6 gives the presentation with a^2=b^3

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).

Inside $G\ast_KH$, the images of $G$ and $H$ intersect exactly in their common image of $K$. (The two factor images intersect exactly in the amalgamated subgroup).

Every subgroup $H$ of a cyclic group $G=\langle g\rangle$ is cyclic. If $H\ne\{e\}$, then the least positive integer $d$ for which $g^d\in H$ satisfies $H=\langle g^d\rangle$. (Every subgroup of a cyclic group is cyclic; the least positive exponent in a nontrivial subgroup supplies a generator).