Normal and self-centralizing implies normality-large

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., self-centralizing normal subgroup) must also satisfy the second subgroup property (i.e., normality-large normal subgroup)
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Any Normal subgroup (?) of a group that is self-centralizing, is also normality-large: its intersection with every nontrivial normal subgroup is nontrivial.


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.