Fully invariant subgroup of additive group of Lie ring is derivation-invariant and fully invariant

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Statement

Suppose L is a Lie ring. Suppose S is a Fully invariant subgroup (?) of the additive group of L. Then:

  1. S is a Lie subring of L, i.e., S is closed under the Lie ring operations of L.
  2. S is a Derivation-invariant Lie subring (?) of L, i.e., any derivation of L sends S to itself.
  3. S is a Fully invariant Lie subring (?) of L: any endomorphism of L as a Lie ring sends S to itself.

Related facts

Similar facts

Applications

Proof

(Using notation as in the statement above).

Since S is a subgroup of L by definition, proving (1) and (2) only requires us to show that S is closed under arbitrary derivations (this would imply both (1) and (2) since the Lie bracket with an element of S is itself an inner derivation). This, in turn, follows from the fact that any derivation is an endomorphism of the underlying abelian group structure, and S, being fully invariant, is thus sent to itself by this endomorphism.

For (3), note that any endomorphism of L as a Lie ring is also an endomorphism of the underlying abelian group structure of L. Since S is fully invariant, the endomorphism must send S to within itself.