3-subnormal subgroup need not have a unique fastest ascending subnormal series

Statement

Suppose $B$ is a group and $A$ is a 3-subnormal subgroup (?) of $B$. Then, it is not necessary that $A$ have a fastest ascending Subnormal series (?) in $B$.

The existence of a fastest ascending subnormal series could fail in two ways:

• There is no unique largest 2-subnormal subgroup (?) of $B$ among those that normalize $A$.
• There is a unique largest 2-subnormal subgroup (?), say $C$ of $B$ among those that normalize $A$. However, there is a smaller subgroup that is part of another ascending subnormal series that overtakes the fastest one involving $C$ at the next stage.

Related facts

• 2-subnormal subgroup has a unique fastest ascending subnormal series
• Subnormal subgroup has a unique fastest descending subnormal series

Facts used

1. Normality is not transitive: This is important to have examples of 2-subnormal subgroups that are not normal.
2. 2-subnormality is not finite-join-closed: A join of two 2-subnormal subgroups need not be 2-subnormal.
3. Subnormality satisfies image condition
4. Subnormality satisfies inverse image condition

Proof

Both proofs rely on the same construction, and liberally use facts (3) and (4), which together yield that the subnormal depth of the full inverse image of a subnormal subgroup under a surjective homomorphism equals the subnormal depth of the subgroup itself. In symbols, if $N$ is normal in $M$, and $N\leq P\leq M$, the subnormal depth of $P$ in $M$ equals the subnormal depth of $P/N$ in $M/N$.

Example of a 3-subnormal subgroup for which there are two ascending subnormal series with one lagging but overtaking

Let $H$ be a nontrivial 2-subnormal subgroup of $G$ that is not normal. Say the fastest ascending subnormal series for $H$ is:

$H<K<G$.

Let $p$ be a prime. Let $A$ be the wreath product of the group of prime order with $G$. Equivalently, $A$ is the semidirect product of the additive group of $\mathbb {F} _{p}[G]$ with $G$ acting by left multiplication.

Let $B=\mathbb {F} _{p}[H]$. Then, we have two subnormal series for $B$ in $A$:

$B\leq \mathbb {F} _{p}[G]\rtimes H\leq \mathbb {F} _{p}[G]\rtimes K\leq A$

and

$B\leq \mathbb {F} _{p}[G]\leq A\leq A$.

Note that if there does exist a fastest ascending subnormal series, it must be the first, because $\mathbb {F} _{p}[G]\rtimes H$ is the unique largest 2-subnormal subgroup of $A$ that is in $N_{A}(B)$ (in fact, it is equal to the whole group $N_{A}(B)$). However, the second term of the second series is bigger than the second term of the first series, so there can be no fastest ascending subnormal series.

Example of a 3-subnormal subgroup with no unique largest normalizing 2-subnormal subgroup containing it

Let $H$ be a nontrivial subgroup of a finite group $G$ that is a join of two 2-subnormal subgroups $H_{1},H_{2}$ of $G$ but is not itself 2-subnormal (such subgroups exist, see fact (2)). In symbols, $H=\langle H_{1},H_{2}\rangle$.

Let $p$ be a prime. Let $A$ be the wreath product of the group of prime order with $G$. Equivalently, $A$ is the semidirect product of the additive group of $\mathbb {F} _{p}[G]$ with $G$ acting by left multiplication.

Let $B=\mathbb {F} _{p}[H]$. Then, $N_{A}(B)=\mathbb {F} _{p}[G]\rtimes H$. This is, by assumption, not 2-subnormal. However, any 2-subnormal subgroup that normalizes $B$ and contains $B$ must be in this. Further, we can see that the subgroups:

$C_{1}=\mathbb {F} _{p}[H]\rtimes H_{1},C_{2}=\mathbb {F} _{p}[H]\rtimes H_{2}$

both are in $N_{A}(B)$ and 2-subnormal. Hence, if there is a unique largest 2-subnormal subgroup of $A$ in $N_{A}(B)$ it must contain both $C_{1}$ and $C_{2}$. But $\langle C_{1},C_{2}\rangle =N_{A}(B)$, which is not 2-subnormal by assumption and fact (3).