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You can use covering spaces instead of cohomology.

(Edit: This proof does not use a presentation of the group, either -- just the weaker fact that its abelianization is free of rank $2g$.)

As Daniel pointed out in his comment, if the group were free it would have to be free on $2g$ generators. The genus $g$ surface has a $2$-sheeted covering space which is a genus $2g-1$ surface. Every index $2$ subgroup of a free group on $r$ generators is free on $2r-1$ generators, because it is the fundamental group of a $2$-sheeted covering space of a wedge of $r$ circles. So $2(2g-1)=2(2g)-1$, a contradiction.

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You can use covering spaces instead of cohomology.

As Daniel pointed out in his comment, if the group were free it would have to be free on $2g$ generators. The genus $g$ surface has a $2$-sheeted covering space which is a genus $2g-1$ surface. Every index $2$ subgroup of a free group on $r$ generators is free on $2r-1$ generators, because it is the fundamental group of a $2$-sheeted covering space of a wedge of $r$ circles. So $2(2g-1)=2(2g)-1$, a contradiction.