To construct the fundamental group of a triangle with edges identified, say ab, bc and ca, I considered the interior $U$ which is contractible and $V$ a thin neibourhood of the perimeter which can be deformation retracted into the three edges which are all the same and is equivalent to $S^1$. The intersection is also equivalent to a circle.
Hence, by van Kampen's theorem, the fundamental group of the triangle, say $X$, $\pi_1(X)=\mathbb{Z}/N$ where $N$ is the normal subgroup (correct me if I got the fundamental group wrong).
What I am not sure here is the normal subgroup. What is this $N$ and how should I express the $N$?
You already have that $\pi_1(U \cap V) \cong \mathbb{Z}$, and you know that $U$ is contractible so you only need to see what your map $\pi_1(U \cap V) \rightarrow \pi_1(V)$ is.
You already said that $V$ deformation retracts to the edges which are in fact a circle. Namely the generator of $\pi_1(V)$ is given by the circle represented by one edge.
It should be clear that the map $\pi_1(U \cap V) \rightarrow \pi_1(V)$ simply sends the generator corresponding to $U \cap V$ to the 3 times the generator corresponding to $V$. (Namely the circle given by the intersection clearly deforms onto the three edges) So your normal subgroup is $3\mathbb{Z}$ and hence your $\pi_1$ for the triangle with edges identified is $\mathbb{Z}/3\mathbb{Z}$.