= Solution
Let $m$ be the <order of a group element> $g$, allowing $m=\infty$. For finite $m$, take $C=\langle a\mid a^{nm}=1\rangle$; for infinite $m$, take $C=\langle a\mid\ \rangle\cong\mathbb Z$. Set
$$
G_1=G*C.
$$
The <normal form theorem for a free product> embeds both factors, so $G$ is a <free factor>. The cyclic subgroup $A=\langle a^n\rangle$ has order $m$, just as $D=\langle g\rangle$ does. Hence $a^{nr}\mapsto g^r$ is a well-defined <group isomorphism> $A\to D$, and we may form the <HNN extension>
$$
H=\langle G_1,t\mid t^{-1}a^nt=g\rangle.
$$
The single displayed conjugacy relation implements the full cyclic isomorphism by taking integer powers. <Britton's lemma> embeds $G_1$ in $H$, so the original $G$ is preserved. Now take
$$
\boxed{h=t^{-1}at.}
$$
Cancellation of successive $tt^{-1}$ gives $h^n=t^{-1}a^nt=g$. This proves the <adjoining a root by an HNN extension> construction and all required embedding claims. If $g=1$, then $m=1$, $C=C_n$, and $A=D=\{1\}$; the HNN relation is vacuous and the same calculation gives $h^n=1$. The construction also works for $n=1$. It uses the actual order of $g$ as an existence parameter, not an algorithm for determining an unknown order.
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