I had a question regarding the cryptogram of Schoenberg's famous hexachord. The Schoenberg hexachord has a pitch class set of (012569) and a vector of <3,1,3,4,3,1>. In other words, the notes are C D-flat D-natural F G-flat A. How does this decipher into spelling out the anagram "A. SCHBEG" for "A. SCHoenBErG"? I roughly know how the French and German cryptograms are done. In German, all letters of notes are equal to letters in the alphabet except for B-natural is H. B-flat is B. I also know that E-flat is Es (S).

Knowing this, I still can't figure how to decipher this cryptogram. Lastly, The complement of this hexachord is supposedly C D-flat E-flat E-natural G A-flat. Why is this so? I thought complements were pitch class sets that filled in the remaining notes of the twelve tone scale that the previous pitch class set did not include. This doesn't seem to be the case here, though.

1 Answer 1


To decode it you have to understand the concept of the set itself and how it can be manipulated. The set is given that nickname due to one of its transpositions and the German name for the notes of that set. If you start the Schoenberg's Hexachord on B♭ you end up with B♭, B, C, E♭, E, and G. Now rewriting it using the German name for some of the notes, you have B, H, C, Es, E, and G which rearranged gives you Schbeg which the only letter missing are oenr.

More information can be found here on the Wiki about it which even has the history of its use in music.

  • That doesn't seem to be said hexachord. I think it's actually supposed to be laid out B♭ B C E♭ E G, yielding only SCHBEG. — Anyway... this is so arbitrary (like everything in twelve-tone technique, mind); you could read plenty of other stuff into that hexachord. Mar 29, 2016 at 22:17
  • @leftaroundabout most thing in music are arbitrary when you get down to a certain level, but you are right the A is not part of the hexachord and that has been fixed. The A does fit in somehow, but it's been too long since I've done set theory that I forget how it fits in.
    – Dom
    Mar 29, 2016 at 22:39
  • Thank you so much, everyone. I tried transposing the set up a sixth because I thought the only way to have the A. in "A. Schoenberg" would be to start on A. Never tried starting on B-flat. I know it doesn't matter much, but it was just one of those things that wasn't clicking for me, and all after having a very confusing lecture on interval vectors from my Professor. Again, thank you.
    – Kalander
    Mar 30, 2016 at 0:56

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