Geometric interpretation of the Fundamental group being non-abelian What topological information a fundamental group of a space based at a point carries?

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*It contains information about the number of one dimensional holes in the space.

What other information does it carry?
Do the fundamental group being abelian or non-abelian gives some topological information about the space?
 A: There is something to be said about the $2$-cells in the $CW$-structure of the path-component containing the basepoint. Namely, that they include the relations of the commutator subgroup of the fundamental group iff the fundamental group is abelian. I will explain what I mean in more detail. Note that if the space in question is not a $CW$-complex, one could say something about its $CW$-approximations.
Recall that the fundamental group of a $CW$-complex only depends on the $2$-skeleton, so we can really only get information that only depends on the $2$-skeleton. Futhermore, we can also only get information about the path-component of the basepoint.
Let $G$ be a group. Take a presentation of $G$ described as a short exact sequence $$0\to K \to F \to G \to 0$$ where $F$ is a free group and the relations of $G$ are given by $K$. Then $G \cong F/K$ is abelian iff $K$ contains the commutator subgroup of $F$.
Let $X$ be a CW-complex. We assume that $X$ is path-connected since we can only say something about the path-component with the basepoint. The $1$-skeleton $X_1$ is homotopy equivalent to a wedge of circles $X_1 \simeq \bigvee_{\alpha\in \mathcal{A}} S^1_\alpha$. These circles correspond to our free generators of $F$ in the short exact sequence and $\pi_1(X_1)\cong F$. The $2$-cells glued to $X_1$ correspond to the relations given in $K$ such that one obtains $\pi_1(X_2)\cong G$. Thus, $\pi_1(X)$ is abelian iff the $2$-cells contains relations generating the commutator subgroup. This what was I meant with the statement.
Let's take a concrete example. Let $T$ denote the torus $S^1\times S^1$. We know that $\pi_1(T)\cong \mathbb{Z} \oplus \mathbb{Z}$, so $T$ much have some $2$-cells generating the commutator subgroup. The usual example of a $CW$-structure for $T$ is given by one $0$-cell, two $1$-cells, and one $2$-cell, so the $1$-skeleton is a wedge of two circles. The circles correspond to the two generators of the fundamental group of $T$. Let's denote the generators by $\alpha$ and $\beta$. Then the $2$-cell is glued to the $1$-skeleton according to the relation $\alpha\beta\alpha^{-1}\beta^{-1}$. Note that $\alpha\beta\alpha^{-1}\beta^{-1}$ is the commutator of the two generators and generates the commutator subgroup as claimed.
