If you could verify my solution, that would be great.
Compute the colimit of the following diagram. \begin{array}{c c c} \langle x \rangle & \xrightarrow{} & D_{n} \\ \downarrow & & \\ 1 & & \end{array} Here $D_n = \langle x,y\colon x^{n} = y^2 = 1, yxy^{-1} = x^{-1}\rangle$ and $n \ge 2$.
I claim that the colimit is $Z_2$ together with the homomorphisms $id: 1 \to Z_2$ and $\varphi: D_n \to Z_2$ where $\varphi(x) = 1$, $\varphi(y) = z$, and $\varphi(x^my^k) = \varphi(x)^m\varphi(y)^k$. Here, $Z_2 = \{e, z\}$ is the group of two elements. Any cocones is precisely the data $L,f: 1 \to L,g: D_n \to L$ of the diagram, where we must have $g(x) = g \circ inc (x) = 1 \circ f = 1$ since $f$ is a homomorphism and thus sends $1$ to $e$. Thus $Z_2, id, \varphi$ is a cone. Also, given $L,f : 1 \to L, g: D_n \to L$, the map $h_L : Z_2 \to L$ given by $h_L(z) = g(y)$ and obviously $h_L(e) = 1_L$ satisfy $h \circ id (1) = f(1) = 1$ and for $x \in D_n$ we have $h \circ \varphi (x) = 1 = g(x) = 1$ and $h \circ \varphi (y) = h(z) = g(y)$, so that $h$ is the corresponding element for the cocone $L$ under the Yoneda perspective. Conversely, it's easy to see that any map $h$ from $Z_2$ to $L$ that would make the diagram commute would have to satisfy $h(e) = 1_L$ and $h(z) = g(y)$. Thus $Z_2, id, \varphi$ is the colimit.
Is my work above correct?