Equivalence of definitions of intermediately subnormal-to-normal subgroup

The definitions that we have to prove as equivalent
The following are equivalent for a subgroup $$H$$ of a group $$G$$:


 * 1) For any intermediate subgroup $$K$$ such that $$H$$ is a subnormal subgroup of $$K$$, $$H$$ is actually a normal subgroup of $$K$$.
 * 2) For any intermediate subgroup $$L$$ such that $$H$$ is a 2-subnormal subgroup of $$L$$, $$H$$ is actually a normal subgroup of $$L$$.
 * 3) For any intermediate subgroup $$M$$, $$N_M(N_M(H)) = N_M(H)$$.

Facts used

 * 1) uses::Normalizer of intermediately subnormal-to-normal implies self-normalizing
 * 2) uses::Normality is strongly join-closed
 * 3) uses::Normality satisfies inverse image condition

(1) implies (2)
This is obvious, since any 2-subnormal subgroup is also subnormal.

(2) implies (1)
Given: A group $$G$$. A subgroup $$H$$ that is normal in any intermediate subgroup $$L$$ in which it is 2-subnormal.

To prove: $$H$$ is normal in any intermediate subgroup $$K$$ in which it is normal.

Proof: We prove this by induction on the subnormal depth of $$H$$, showing that if $$H$$ has subnormal depth $$r$$ in $$K$$ it has subnormal depth $$r - 1$$, for $$r \ge 2$$. Indeed, suppose $$H$$ has subnormal depth $$r$$. We have a subnormal series:

$$H = H_0 \le H_1 \le H_2 \le \dots \le H_r = K$$.

Then, $$H_0$$ is 2-subnormal in $$H_2$$, hence is normal in $$H_2$$ by assumption. Thus, we hav ea subnormal series of length $$r - 1$$:

$$H = H_0 \le H_2 \le \dots \le H_r = K$$.

This completes the induction.

(2) implies (3)
This follows from fact (1). Note that fact (1) only states that the normalizer in the whole group is self-normalizing, but since the property of being intermediately subnormal-to-normal is preserved on looking at intermediate subgroups, the normalizer in any intermediate subgroup is self-normalizing.

(3) implies (2)
Given: A group $$G$$, a subgroup $$H$$ such that $$N_M(N_M(H)) = N_M(H)$$ for any intermediate subgroup $$M$$. $$H$$ is 2-subnormal in some intermediate subgroup $$L$$

To prove: $$H$$ is also normal in $$L$$.

Proof: Let $$P = N_L(H)$$. We prove that $$P = L$$, by deriving a contradiction from the assumption that $$P \ne L$$.


 * 1) Let $$N$$ be the join of all normal subgroups of $$L$$ containing $$H$$ and contained in $$P = N_L(H)$$. Note that the set is nonempty, since $$H$$ is $$2$$-subnormal in $$L$$. By fact (2), $$N$$ is itself normal in $$L$$.
 * 2) Suppose $$P \ne L$$. Let $$x \in L \setminus P$$. Then, define $$M = \langle N, x \rangle$$ and $$Q = M \cap P$$. Then,
 * 3) $$N_M(H) = N_L(H) \cap M = P \cap M = Q$$.
 * 4) Also, $$Q/N$$ is a subgroup of the cyclic group $$M/N$$, and is hence normal in $$M/N$$. Thus, by fact (3), $$Q$$ is normal in $$M$$. Thus, $$N_M(Q) = M$$.
 * 5) Thus, $$N_M(N_M(H)) = M$$, while $$N_M(H) = Q = M \cap P$$. Since by construction, $$M$$ is not contained in $$P$$, we obtain that $$N_M(N_M(H)) \ne N_M(H)$$, and we have a contradiction.