Based on the English alphabetical order, three of the following four…

A letter-cluster classification is decided on alphabet positions, not on how the letters look. Replace every letter by its position in the English alphabet (A = 1, B = 2, ... Z = 26); the group is then defined by two positional properties that every member must satisfy: a fixed step between the two letters inside each half of the pair, and a fixed shift that carries the first half onto the second, with the count continuing cyclically from Z back to A. The member that fails those properties is the one outside the group.

Written in alphabet positions, the four pairs are:

Comparing the rows:

• VX-BD, GI-MO and UW-AC each keep a +2 step inside both halves and a uniform +6 shift from the first half onto the second.

• JL-OT keeps the +2 step only in its first half; O to T is a +5 step, and its two shifts, J to O (+5) and L to T (+8), are unequal, so it matches neither property.

• The cyclic count is what lets the wrapping pairs follow the same rule: U(21) + 6 = 27 -> 1 = A and W(23) + 6 = 29 -> 3 = C, exactly as G(7) + 6 = 13 = M.

• This also agrees with the note in the question: the three grouped pairs do not share a common vowel or consonant count (VX-BD has no vowel, GI-MO has two, UW-AC has two), so the rule has to be positional.

Therefore the pair that does not belong to the group is JL-OT.