Normal and self-centralizing implies normality-large

Statement
Any fact about::normal subgroup of a group that is self-centralizing, is also normality-large: its intersection with every nontrivial normal subgroup is nontrivial.

Proof
Given: A group $$G$$, a normal subgroup $$N$$, such that $$C_G(N) \le N$$.

To prove: If $$M$$ is a nontrivial normal subgroup of $$G$$ such that $$M \cap N$$ is trivial, then $$M$$ is trivial.

Proof: Consider the commutator $$[N,M]$$. This is contained both in $$N$$ and in $$M$$, since they are both normal. Hence it is contained in $$N \cap M$$, which is trivial. Thus, every element of $$N$$ commutes with every element of $$M$$, so $$M \le C_G(N)$$. Since $$N$$ is self-centralizing, we get $$M \le N$$. Since $$N \cap M$$ is trivial, we get that $$M$$ is trivial.