Lie ring is abelian iff every subring is an ideal

Statement
Suppose $$L$$ is a Lie ring. The following are equivalent for $$L$$:


 * 1) $$L$$ is an abelian Lie ring, i.e., the Lie bracket of any two elements of $$L$$ is zero.
 * 2) Every subring of $$L$$ is an ideal of $$L$$.

Failure of analogous statement for groups
The analogous statement is false for groups. Explicitly, a group in which every subgroup is normal is termed a Dedekind group, and there do exist non-abelian Dedekind groups, called Hamiltonian groups. The smallest example is the quaternion group of order eight.

(1) implies (2)
This is obvious.

Proof outline for (2) implies (1)
We prove by contradiction. Consider a non-abelian Lie ring $$L$$ satisfying condition (2). Start with $$x,y \in L$$ such that $$[x,y] \ne 0$$. Note that $$\langle x,y \rangle$$ is therefore not cyclic. By a version of the structure theorem for finitely generated abelian groups, we can find from these two new elements $$u,v$$ such that the cyclic subgroups generated by $$u$$ and $$v$$ intersect trivially and $$[u,v] \ne 0$$.

Now, both the subgroups $$\langle u \rangle$$ and $$\langle v \rangle$$ are subrings. Therefore, by (2), they are ideals. This means that $$[u,v] \in \langle u \rangle$$ and $$[u,v] \in \langle v \rangle$$, so $$[u,v] = 0$$, contradicting the assumption.