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I had an example in my topology script that the hurewicz homomorphism factors over the abelianization of the fundamental group. How is this useful?

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    It is a bit stronger than that, saying that for many spaces, the homomorphism _is_ the abelianisation of the fundamental group. This _is_ useful. For instance, it greatly reduces the possible [knot groups](https://en.wikipedia.org/wiki/Knot_group). – 2017-01-23

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If your space is path connected, then the map $$\bar{h}:\frac{\pi_1(X,x_0)}{[\pi_1(X,x_0),\pi_1(X,x_0)]}\to H_1(X)$$ which the Hurewicz map $h:\pi_1(X,x_0)\to H_1(X)$ factors through is an isomorphism. This is helpful if you know $\pi_1$ and want to compute $H_1$.

For example, if we already know that $\pi_1(S^1\vee S^1,s_0) = \mathbb{Z}\ast\mathbb{Z}$, the free group on two generators, then by the above isomorphism it follows that $H_1(S^1\vee S^1)=\mathbb{Z}\oplus\mathbb{Z}$, since the abelainization of the free group on $n$ generators is the free abelian group on $n$ generators.

Similarly, if $K$ is the Klein bottle, we can use the CW decomposition to show that $$\pi_1(K,k_0)\cong \langle a,b\ |\ abab^{-1}\rangle.$$ The abelianization of the above is just $$H_1(K) = \langle a,b\ |\ abab^{-1},ab=ba\rangle = \langle a,b\ |\ a^2,ab=ba \rangle$$ since $ab=ba$ implies that $abab^{-1}=a^2$. It shouldn't be too hard to believe that the above group is just $\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$.