How to determine space with a given fundamental group. I would like to give examples of topological spaces such that their fundamental groups are, respectively, $\mathbb{Z} \oplus \mathbb{Z}_{n}$ and $\mathbb{Z} \ast \mathbb{Z}_{n}$. For the latter, I thought about this: let $X_{n}$ be the space obtained from $S^{1}$ by attaching a 2-cell with characteristic map $\chi (z) = z^{n}$. This map identifies all $n$-th roots of unity and can therefore be described as a quotient of $D^{2}$ with labels $a\ldots a$ ($n$ times). It's fundamental group, let alone, is $\langle a | a^{n} \rangle$, which is $\mathbb{Z}_{n}$ (in particular, if $n=2$, $X_{n} = P^{2}$). Now I thought about wedge-summing a copy of $S^{1}$ with this space. The resulting space would be, then, a space with fundamental group $\mathbb{Z} \ast \mathbb{Z}_{n}$, by Seifert Van Kampen's theorem. Is that right? Now, for the second part, I'm completely lost. If possible, I'd like to avoid using that the fundamental group of the cartesian product is the direct sum of fundamental groups.
 A: Your construction of a space with $\pi_1\cong\mathbb{Z}\ast\mathbb{Z}_n$ is good and the standard way to construct such a space.
The standard way to find an $X$ such that $\pi_1(X)\cong\mathbb{Z}\oplus\mathbb{Z}_n$ is, as you hinted at, to simply find a space $X_1$ with $\pi_1(X_1)\cong \mathbb{Z}$ and $X_2$ with $\pi_1(X_2)\cong \mathbb{Z}_n$ and then use the fact that $\pi_1(X_1\times X_2)\cong \pi_1(X_1)\times\pi_1(X_2)$. Using this approach, we find that the space $S^1\times (D^2\cup_{z^n}S^1)$ has the necessary fundamental group (here $(D^2\cup_{z^n}S^1)$ is the space you described in your question).
We can construct such a space though, without resorting to using the above theorem of product spaces. Note that the fundamental group of the torus $T^2$ is $\mathbb{Z}\times\mathbb{Z}$ which is generated by the meridinal and longitudinal loops on the torus. Call these representative elements $\alpha$ and $\beta$ respectively. Similarly to how you glued a disk to a circle to kill off certain elements in the fundamental group, we can do the same here. If we glue the boundary of $D^2$ to the image of the loop $\alpha$ via the map $f\colon z\mapsto z^n$ as before, then we introduce the relation $\alpha^n=0$ in the fundamental group, and as $\pi_1(T^2)$ is free abelian on the generators $\alpha,\beta$, there is no relation introduced on to the generator $\beta$. Hence $\pi_1(T^2\cup_fD^2)\cong \mathbb{Z}_n\oplus\mathbb{Z}$.
You can prove this more rigorously using Van-Kampen's theorem (take a small open neighbourhood on the torus around the loop $\alpha$, union the disk as the set $U$, and the torus minus the disk (and its boundary) as the set $V$).
