2 added 11 characters in body

Perhaps it isn't necessary to add another answer, but here is one more nevertheless.

1. Given any discrete group $G$, form the classifying space $BG$. A map of $S^1$ to $BG$ is equivalent to giving a principal $G$-bundle over $S^1$. Such a bundle can formed by gluing the ends of the trivial bundle $[0,1]\times G$ with a twist given by an element of $G$. So in this way, we get $\pi_1(BG)=G$.

2. Of course, invoking $BG$ is perhaps violating the spirit of the question. When $G$ is finite, we can approximate this by something more concrete. Let $V$ be a faithful unitary representation of $G$ with no trivial summands (e.g. the complement of $\mathbb{C}$ in the regular representation). After replacing $V$ by $V\oplus V$ if necessary, we can assume that $\dim V>1$. Then the unit sphere $S\subset V$ will be simply connected, and $G$ will act on it without fixed points. So the quotient $S/G$ will have fundamental group $G$. This is the same construction used (implicitly) in many of the other answers.

1

Perhaps it isn't necessary to add another answer, but here is one more nevertheless.

1. Given any group $G$, form the classifying space $BG$. A map of $S^1$ to $BG$ is equivalent to giving a principal $G$-bundle over $S^1$. Such a bundle can formed by gluing the ends of the trivial bundle $[0,1]\times G$ with a twist given by an element of $G$. So in this way, we get $\pi_1(BG)=G$.

2. Of course, invoking $BG$ is perhaps violating the spirit of the question. When $G$ is finite, we can approximate this by something more concrete. Let $V$ be a faithful unitary representation of $G$ with no trivial summands (e.g. the complement of $\mathbb{C}$ in the regular representation). After replacing $V$ by $V\oplus V$ if necessary, we can assume that $\dim V>1$. Then the unit sphere $S\subset V$ will be simply connected, and $G$ will act on it without fixed points. So the quotient $S/G$ will have fundamental group $G$. This is the same construction used (implicitly) in many of the other answers.