Is there a subgroup of order 2^n (for large n) of the prime field that secp256k1 is defined over?

No, there cannot be.

There are two groups defined by the secp256k1 field, the addition group, and the group of multiplication of its non-zero elements.

Neither has a large 2ⁿ subgroup, because the order of a subgroup of a finite group must be a divisor of the order of that group.

• The order of the additive group must divide the size of the field, namely p, the modulus itself, which is prime, and thus only has itself and the trivial group as subgroups.

• The order of the multiplicative group of a field is one less than the size of the field, so p-1, which factors as 2 * 3 * 7 * 13441 * 205115282021455665897114700593932402728804164701536103180137503955397371. Thus, it only has C₂ (the cyclic group with two elements) as subgroup whose order is a power of two. Specifically, in the embedding in the secp256k1 multiplicative group, that subgroup consists of the elements 1 and p-1, and the multiplication between them.

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