Equivalence of definitions of intermediately subnormal-to-normal subgroup

This article gives a proof/explanation of the equivalence of multiple definitions for the term intermediately subnormal-to-normal subgroup
View a complete list of pages giving proofs of equivalence of definitions

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. Normalizer of intermediately subnormal-to-normal implies self-normalizing
2. Normality is strongly join-closed
3. Normality satisfies inverse image condition

Proof

(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 $r\geq 2$. Indeed, suppose $H$ has subnormal depth $r$. We have a subnormal series:

$H=H_{0}\leq H_{1}\leq H_{2}\leq \dots \leq H_{r}=K$.

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 $r-1$:

$H=H_{0}\leq H_{2}\leq \dots \leq H_{r}=K$.

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 $P\neq L$.

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 $P\neq L$. Let $x\in L\setminus P$. Then, define $M=\langle N,x\rangle$ and $Q=M\cap P$. Then,
   1. $N_{M}(H)=N_{L}(H)\cap M=P\cap M=Q$.
   2. 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, $N_{M}(Q)=M$.
   3. 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 $N_{M}(N_{M}(H))\neq N_{M}(H)$, and we have a contradiction.