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There are many good answers already, but maybe you'll be interested in a counting argument.

Let $Q_8$ denote the quaternion group. Consider the set

$$\mbox{Hom}(\pi_1(X),Q_8)$$

where $X$ is the closed surface of genus $g$. Since we have a presentation for $\pi_1(X)$, we note that a homomorphism to $Q_8$ is given by a $2g$-tuple of elements in $Q_8$ satisfying some group word. In particular, the cardinality of the set is finite. Indeed, if the fundamental group were free, the cardinality would be a power of $8$. However, combinatorics (or gentle representation theory) gives the cardinality of the set exactly:

$$|\mbox{Hom}(\pi_1(X),Q_8)| = 2^{6g-1}+2^{4g-1}$$

which (for $g>0$) is never a power of $2$, still less a power of $8$.

---Edit---

The "gentle representation theory" to which I refer can be found here: http://arxiv.org/abs/1102.4353

We show in section 4 that the cardinality of $\mbox{Hom}(\pi_1(X),G)$ can be written explicitly in terms of the dimensions of the irreducible representations of $G$.

1

There are many good answers already, but maybe you'll be interested in a counting argument.

Let $Q_8$ denote the quaternion group. Consider the set

$$\mbox{Hom}(\pi_1(X),Q_8)$$

where $X$ is the closed surface of genus $g$. Since we have a presentation for $\pi_1(X)$, we note that a homomorphism to $Q_8$ is given by a $2g$-tuple of elements in $Q_8$ satisfying some group word. In particular, the cardinality of the set is finite. Indeed, if the fundamental group were free, the cardinality would be a power of $8$. However, combinatorics (or gentle representation theory) gives the cardinality of the set exactly:

$$|\mbox{Hom}(\pi_1(X),Q_8)| = 2^{6g-1}+2^{4g-1}$$

which (for $g>0$) is never a power of $2$, still less a power of $8$.