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Original proof. How about a surface group is 1-ended since its universal cover is the hyperbolic plane and the non-abelian free group has infinitely many ends since the universal cover of the wedge of circles is a tree.

Second proof. Since Misha has suggested a proof from each area here is a proof rom from universal algebra. Let V be the variety of extensions of elementary abelian 2-groups by elementary abelian 2-groups. So commutators commute. The word problem for the relatively free group on X in this variety is well-known. A word is trivial if and ony if it is trivial in $(\mathbb Z/2)^X$ and the loop it labels in the Cayley graph of $(\mathbb Z/2)^X$ with respect to $X$ traverses each geometric edge an even number of times.

If the surface group were free it would have to be free on $2g$-generators because of the abelianization. So if we factor by the verbal subgroup associated to V we would get a free group in this variety on 2-generators. But the group in V on 2g-generators with the surface defining relation is a proper quotient of the relatively free group because the product of commutators in question uses each edge of the Cayley graph of $(\mathbb Z/2)^{2g}$ exactly once. Since these are finite groups, a proper quotient is not free.

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Original proof. How about a surface group is 1-ended since its universal cover is the hyperbolic plane and the non-abelian free group has infinitely many ends since the universal cover of the wedge of circles is a tree.

Second proof. Since Misha has suggested a proof from each area here is a proof rom universal algebra. Let V be the variety of extensions of elementary abelian 2-groups by elementary abelian 2-groups. So commutators commute. The word problem for the relatively free group on X in this variety is well-known. A word is trivial if and ony if it is trivial in $(\mathbb Z/2)^X$ and the loop it labels in the Cayley graph of $(\mathbb Z/2)^X$ with respect to $X$ traverses each geometric edge an even number of times.

If the surface group were free it would have to be free on $2g$-generators because of the abelianization. So if we factor by the verbal subgroup associated to V we would get a free group in this variety on 2-generators. But the group in V on 2g-generators with the surface defining relation is a proper quotient of the relatively free group because the product of commutators in question uses each edge of the Cayley graph of $(\mathbb Z/2)^{2g}$ exactly once. Since these are finite groups, a proper quotient is not free.

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How about a surface group is 1-ended since its universal cover is the hyperbolic plane and the non-abelian free group has infinitely many ends since the universal cover of the wedge of circles is a tree.