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One can find generators for the ring of invariants of $\mathbb F_2[x_{ij}:1\leq i < j \leq n]$ under the action of $S_n$ on the indices, which are finitely many by Noether's finite generation theorem. I think this gives you a complete set of invariants.

Later: As Steve observes, one would like the number of invariants not to grow too fast. In characteristic zero (which we may use just as well), Noether's bound tells us that the ring of invariants is generated by at most $\binom{n^2+n!}{n!}$ elements, but this is quite huge (for $n=6$ the bound is 48813025503084826957958990535221725233495346780817632847728425, which is discouraging...) I do not think anyone knows how many elements one really needs, though, to generate---usually generate in this particular case---usually Noether's bound is pretty bad.

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One can find generators for the ring of invariants of $\mathbb F_2[x_{ij}:1\leq i < j \leq n]$ under the action of $S_n$ on the indices, which are finitely many by Noether's finite generation theorem. I think this gives you a complete set of invariants.

Later: As Steve observes, one would like the number of invariants not to grow too fast. In characteristic zero (which we may use just as well), Noether's bound tells us that the ring of invariants is generated by at most $\binom{n^2+n!}{n!}$ elements, but this is quite huge (for $n=6$ the bound is 48813025503084826957958990535221725233495346780817632847728425, which is discouraging...) I do not think anyone knows how many elements one really needs, though, to generate---usually Noether's bound is pretty bad.

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One can find generators for the ring of invariants of $\mathbb F_2[x_{ij}:1\leq i < j \leq n]$ under the action of $S_n$ on the indices, which are finitely many by Noether's finite generation theorem. I think this gives you a complete set of invariants.