A monomorphism of groups which is not universal?

If $f_i :G \to H_i$ are two homomorphisms, and $H_i = \langle X_i | R_i \rangle$ are group presentations, then the pushout has the group presentation $$H_1 \sqcup_G H_2 = \langle X_1,X_2 : R_1, R_2 , \{f_1(g)=f_2(g)\}_{g \in G} \rangle.$$ Formally, one has to express $f_1(g)$ (likewise $f_2(g)$) here in terms of the generators $X_1$ (resp. $X_2$) so that $f_1(g)=f_2(g)$ becomes, in fact, a relation between the letters of $X_1$ and $X_2$. Now look what happens for $G=\mathbb{Z}$ and $H_2=\mathbb{Z}/n\mathbb{Z}=\langle t : t^n=1 \rangle$ with $f_2$ the canonical projection. Then $f_1 : \mathbb{Z} \to H_1$ corresponds to an element $h \in H_1$ (via $h=f_1(1)$) and we have $$H_1 \sqcup_G H_2 = \langle X_1,t : R_1, t^n=1,h=t \rangle = \langle X_1:R_1,h^n=1 \rangle = H_1 / \langle\langle h^n \rangle\rangle.$$ Now assume that $f_1$ is injective, i.e. that $h \in H_1$ has infinite order. We ask ourselves if the canonical homomorphism $H_2 \to H_1 \sqcup_G H_2$ is also injective. It identifies with $$\mathbb{Z}/n\mathbb{Z} \to H_1/\langle \langle h^n \rangle\rangle,\,[1] \mapsto [h].$$ Hence, the question is equivalent to: Does $[h]$ have order $n$ in $H_1 / \langle\langle h^n \rangle\rangle$? A priori it is only clear that the order divides $n$. By the way, it is a common mistake to deduce the order of the generators from a given group presentation - sometimes such a group turns out to be trivial! And this leads us to a class of counterexamples, namely $H_1$ could be simple, so that $H_1 / \langle \langle h^n \rangle \rangle = 0$ cannot contain $\mathbb{Z}/n\mathbb{Z}$ for $n>1$. An example of an infinite simple group with an element of infinite order is $\mathrm{PSL}_3(\mathbb{R})$ with $$h = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$