 Solution

ID: past-exam-of-the-lie-algebras-and-representation-theory-of-symmetric-groups-course-of-the-university-of-oxford/2009/a/1/c/solution

Polarizing

[x, [x, z]] = 0

gives

[x, [y, z]] + [y, [x, z]] = 0.

Let

T = [x, [y, z]]

. Antisymmetry and this identity give

[y, [z, x]] = T

and

[z, [x, y]] = T

. The Jacobi identity therefore gives

3 T = 0

. Since the characteristic is not three,

T = 0

for every triple. Consequently

[L, [L, L]] = 0, L has nilpotency class at most two .

No division by two was used, so the proof also applies in characteristic two with the usual alternating definition of a Lie bracket.

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