Consider the group $G$ with presentation $$G=\langle a,b:\hspace{0.1cm}a^2b^{-1}a^3b^2=1\rangle.$$ What is the explicit path-connected topological space $X$ with $\pi_1(X)\cong G$?
My attempt: I try to modify the relations so that I can get a space with known relations like the torus but I can't find them. I also did the quotient of an octagon with the labelling scheme $\hspace{0.1cm}a^2b^{-1}a^3b^2$ however (after simplifications of cutting and paste) I can't understand the resulting space.
There is a general procedure for this type of problem, known as taking a presentation complex. The idea is the same as taking a presentation of a group, and realizing at a quotient of a free group, but we do this topologically:
Step 1: Build a wedge of circles, and lable each circle by a generator for a group. Give this an orientation. In you case this will be $S^1 \vee S^1$.
Step 2: Paste in a $2$-cell according to any relations in your group. Pasting in a two cell along the word in $a,b$ kills that element of $\pi_1(S^1 \vee S^1)= \langle a,b | \,\,\,\rangle$. in Particular, you can take an octogon and label each edge according to the gluing rules. If you have $a$ in the word, label the first edge $a$. If you have $a^{-1}$ label it with the opposite orientation of $a$, and so on for every letter in your word.
To show that this has the right fundamental group, you need Van-Kampen's theorem.
A good first example, is the group $\mathbb Z^2=\langle a,b \mid aba^{-1}b^{-1}\rangle$. This is basically the wedge of two circles, but you paste in a square according to this rule. This is basically the presentation for a torus, which has the correct fundamental group.
This space can be taken as a model for the $2$-skeleton of a $K(G,1)$ space for your group $G$. You then paste in higher cells to get rid of higher homotopy groups.