Intermediately normal-to-characteristic of normal implies intermediately subnormal-to-normal

This article describes a computation relating the result of the Composition operator (?) on two known subgroup properties (i.e., Intermediately normal-to-characteristic subgroup (?) and Normal subgroup (?)), to another known subgroup property (i.e., Intermediately subnormal-to-normal subgroup (?))
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Statement

Suppose $H\leq K\leq G$ are such that $H$ is an intermediately normal-to-characteristic subgroup of $K$, and $K$ is normal in $G$. Then, $H$ is intermediately subnormal-to-normal in $K$.

Related facts

Weaker facts

• Intermediately normal-to-characteristic implies intermediately subnormal-to-normal

Converse

It is not in general true that if $H\leq K$ is a subgroup such that whenever $K$ is normal in $G$, $H$ is intermediately subnormal-to-normal in $G$, then $H<math>isintermediatelynormal-to-characteristicin<math>K$. Thus, being intermediately normal-to-characteristic is not the left residual of intermediately subnormal-to-normal by normal.

Facts used

1. Subnormality satisfies intermediate subgroup condition
2. Intermediately normal-to-characteristic implies intermediately subnormal-to-normal
3. Normality satisfies transfer condition: The intersection of a normal subgroup with any subgroup is normal in that subgroup.
4. Characteristic of normal implies normal

Proof

Given: Groups $H\leq K\leq G$ such that $H$ is intermediately normal-to-characteristic in $K$ and $K$ is normal in $G$.

To prove: If $L\leq G$ is such that $H$ is subnormal in $L$, then $L$ is normal in $L$.

Proof: Let $M=K\cap L$.

1. $H$ is characteristic in $M$: By fact (1), $H$ is subnormal in $M$. By fact (2), $H$ is normal in $M$, and hence characteristic in $M$.
2. $M$ is normal in $L$: Since $K$ is normal in $G$, fact (3) yields that $M=K\cap L$ is normal in $L$.
3. $H$ is normal in $L$: This follows from the previous two steps, and fact (4).