Consider $CW$-complex $X$ obtained from $S^1\vee S^1$ by glueing two $2$-cells by the loops $a^5b^{-3}$ and $b^3(ab)^{-2}$. As we can see in Hatcher (p. 142), abelianisation of $\pi_1(X)$ is trivial, so we have $\widetilde H_i(X)=0$ for all $i$. And if $\pi_2(X)=0$, we have that $X$ is $K(\pi,1)$.
My question is: how can one compute $\pi_2(X)$? Computing homotopy groups is hard, what methods may i use?
On
Letting $\widetilde X$ be the universal cover of $X$ we have $$\pi_2(X) \approx \pi_2(\widetilde X) \approx H_2(\widetilde X,\mathbb{Z}) $$ The first isomorphism comes from the long exact sequence of homotopy groups of a fibration, using discreteness of the fiber of the universal covering map. The second isomorphism comes from the Hurewicz theorem, using simple connectivity of $\widetilde X$.
No, $\pi_2(X)=\mathbb Z^{119}$, is is obvious by considering Euler characteristic.
Denote universal cover of $X$ by $\widetilde X$. In Hatcher we see (and i don't know how to prove it) that $\pi_1(X)$ has order $120$. So $\widetilde X$ has 120 $0$-cells, 240 $1$-cells, and 240 $2$-cells; therefore $\chi(\widetilde X)=120$.
We know that $H_0(\widetilde X)=\mathbb Z$, $H_1(\widetilde X)=0$ and $H_2(\widetilde X)$ has no torsion. So $H_2(\widetilde X)=\pi_2(\widetilde X)=\pi_2(X)=\mathbb Z^{119}$.