The answer is *yes*, already for an affine variety.

The following example is taken from Dimca's book *Singularities and topology of hypersurfaces*, see page 102 and page 105. We work over $\mathbb{C}$.

Let $V \subset \mathbb{P}^n$ be a hypersurface and $U:=\mathbb{P}^n \setminus V$ its complement. Since $V$ is very ample, $U$ is an affine variety. Then we have the following

> **Proposition.** Assume that $V$ has $k$ irreducible components $V_1, \ldots, V_k$ with $\deg V_i =d_i$. Then $$H_1(U,  \mathbb{Z})=\mathbb{Z}^{k-1} \oplus \mathbb{Z}/d \mathbb{Z},$$
where $d$ denotes the greatest common divisor of the integers $d_1, \ldots, d_k$.

> If moreover $V$ has only normal crossing singularities in codimension $1$, then $\pi_1(U)$ is abelian, hence it is isomorphic to the group $H_1(U, \mathbb{Z})$ given above.

> In particular, if $V$ is irreducible and normal of degree $d$, one has $$\pi_1(U)=\mathbb{Z}/d \mathbb{Z}.$$

**Remark.** If $V$ is irreducible of degree $d$, then the Proposition above implies that in any case the abelianization of $\pi_1(U)$ is isomorphic to $\mathbb{Z}/d \mathbb{Z}$. However, when $V$ is not normal crossing in codimension $1$, the group $\pi_1(U)$ may be nonabelian (and also infinite). Let me give a couple of classical examples, that can be found in Dimca's book, Chapter 4.

*Example 1.* Let $V \subset \mathbb{P}^2$ be the tricuspidal quartic curve of equation $$x^2y^2+y^2z^2+z^2x^2-2xyz(x+y+z)=0.$$ Then $\pi_1(U)$ is isomorphic to the metacyclic group of order $12$. This is a finite nonabelian group, whose abelianization is $\mathbb{Z}/4 \mathbb{Z}$.

*Example 2.* Let $V \subset \mathbb{P}^2$ be the plane sextic of equation $$(x^2+y^2)^3+(y^3+z^3)^2=0.$$ $V$ has six cusps situated on a conic and $\pi_1(U)=(\mathbb{Z}/2 \mathbb{Z}) \ast (\mathbb{Z}/3 \mathbb{Z})$. This is an infinite, nonabelian group whose abelianization is $(\mathbb{Z}/2 \mathbb{Z}) \times (\mathbb{Z}/3 \mathbb{Z})= \mathbb{Z}/6 \mathbb{Z}$.