2-dimesional cell complexes with fundamental group isomorphic to the following. I have been asked to give examples of 2-dimensional cell complexes whose fundamental group isomorphic to the following 
$$ \Bbb Z_4 * \Bbb Z_5$$and 
$$\Bbb Z_4\times \Bbb Z5$$
I know in the first case I am looking for the wedge of 2 spaces whose fundamental groups are $\Bbb Z_4$ and $\Bbb Z_5$ respectively and for the second I'm looking for the the cross product of two spaces with the same fundamental groups.  
However I haven't come across any examples of spaces whose fundamental groups are anything but $\Bbb Z $ either crossed with itself or the free product of $\Bbb Z $ with itself.  
Can anyone give examples of spaces whose fundamental groups are different? 
 A: Here are some hints. Most of what follows is already contained in your question and the comments.
As suggested by Dan in the comments, you should start out by finding two connected $2$-complexes $X$ and $Y$ that have $\Bbb Z/4$ and $\Bbb Z/5$ as fundamental groups. One way of constructing a connected $2$-dimensional CW complex with fundamental group $\Bbb Z/n$ (for some integer $n\geq 2)$ is to start with the circle $S^1$, and to attach a $2$-cell $D^2$ with the attaching map $\partial D^2\to S^1, z\mapsto z^n$.
To obtain the first space you want, just choose base points in $X$ and $Y$ and consider $X\vee Y$. This will give the desired result since the basepoint in $X\vee Y$ has a contractible neighborhood, and using van Kampen's theorem.
For the second construction, you proposed the idea of using $X\times Y$. While this breaks the dimensional requirement, it can be salvaged: just consider the $2$-skeleton of the product complex $X\times Y$. It is a general fact that, for any integers $0\leq k\leq n$, and any CW-complex $X$, the $k$-skeleton inclusion $X_n\hookrightarrow X$ induces isomorphisms on the homotopy groups $\pi_k(X_n)\to\pi_k(X)$ while $k<n$ (and a surjection $\pi_n(X_n)\to\pi_n(X)$). Hence, the $2$-skeleton $Z=(X\times Y)_2$ of the $4$-dimensional product complex $X\times Y$ has its fundamental group isomorphic to that of $X\times Y$, i.e. $\pi_1(Z)\simeq\Bbb Z/4\times\Bbb Z/5$.
Of course, it would be simpler to remember that $\Bbb Z/4\times\Bbb Z/5\simeq\Bbb Z/20$ since $4$ and $5$ are coprime, and apply the first construction.
A: A cell complex with 1 vertex, one edge (both ends attached to that vertex, forming a circle that I'll treat as $S^1$, and one 2-cell, attached to the 1-skeleton via the map 
$$
\theta \mapsto 2\theta
$$
in which the boundary of the 2-cell double-covers the 1-skeleton, has $H_0 = Z$ and $H_1 = Z/2Z$. You should confirm this (and read up on $RP^2$ if you cannot do so). 
Generalize to find a complex with $H_1 = Z/nZ$. 
See what you can do from there. 
Hint: for the second part, you're going to need a generator $a$ for the first cyclic group and a generator $b$ for the second to commute. You can get that by attaching a 2-cell along $aba^{-1}b^{-1}$. If this doesn't make sense, look at the torus and compute its homology to see why the latitude and meridian generators commute. 
