 Solution

ID: past-exam-of-the-lie-algebras-and-representation-theory-of-symmetric-groups-course-of-the-university-of-oxford/2008/a/1/b/solution

Start with

I_{0} = I

and set

I_{k + 1} = [L, I_{k}]

. Every term lies in

I

, and

I_{k} \subseteq L_{k + 1}

, so the sequence eventually vanishes. Choose its last nonzero term

I_{j}

. Then

[L, I_{j}] = 0

, hence

0 \neq = I_{j} \subseteq I \cap Z (L)

. Therefore

I \cap Z (L) \neq = {0}

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