Nilpotent multiplier of abelian group is graded component of free Lie ring

Statement
Suppose $$G$$ is an abelian group and $$c$$ is a positive integer. Denote by $$\mathcal{L}(G)$$ the free Lie ring on $$G$$. Note that $$\mathcal{L}(G)$$ naturally has the structure of a graded Lie ring. The claim is that the $$c$$-nilpotent multiplier $$M^{(c)}(G)$$ (defined as the Baer invariant of $$G$$ with respect to the subvariety of the variety of groups given by groups of nilpotency class at most $$c$$) is isomorphic, as an abelian group, to the $$(c+1)^{th}$$ graded component of $$\mathcal{L}(G)$$:

$$M^{(c)}(G) \cong (c+1)^{th} \mbox{ graded component of } \mathcal{L}(G)$$

Note that the first graded component is $$G$$, so we get that:

$$\mbox{Additive group of } \mathcal{L}(G) \cong G \oplus \bigoplus_{c=1}^\infty M^{(c)}(G)$$

Interpretation in Schur functor terms
The graded components of the free Lie ring of $$G$$ can be viewed as functors of $$G$$. Each of these functors can explicitly described in Schur functor terms: see induction formula for Lie operad.

Related facts

 * Formula for dimension of graded component of free Lie algebra, along with this fact, allows us to explicitly compute the nilpotent multipliers for a free abelian group.
 * Nilpotent multiplier of perfect group equals Schur multiplier

Journal references

 * , Proposition 2.1o (not proved here, but referenced with context)
 * , Proposition 2.1o (not proved here, but referenced with context)