Suppose $n \ge 3$ and $A_1, A_2, \dots, A_n$ are nontrivial groups. Under what conditions can we find a group G and a subgroup H with a chain:

$$H = H_0 \le H_1 \le H_2 \le \dots \le H_n = G$$

such that each $H_{i-1}$ is normal in $H_i$ and $H_i/H_{i-1} \cong A_i$ and such that the subnormal depth (also called the subnormal defect) of H in G is exactly n. In other words, there is no shorter subnormal series for H in G.

Incidentally, this is always possible for $n = 2$. I have a proof here. The key idea is to take the wreath product of the two groups and then take H as "all but one coordinate" in the base normal subgroup of the corresponding semidirect product.

Aside: The wreath product construction probably does not work for $n \ge 3$ because iterating wreath products does not get us beyond a subnormal depth of 2. See this, for instance.


This can always be done:

Given nontrivial groups $A_i$ for $0 \le i \le n$, there exists a group $G$ and a subnormal series $H = H_0 < \cdots < H_n = G$ such that $H_i/H_{i-1} \cong A_i$ for $0 \le i < n$ and such that no shorter subnormal series from $H$ to $G$ exists.

Here is my proof:

We can assume $n > 1$, and we induct on $n$. By the inductive hypothesis, let $W$ be a group with subnormal series $V = V_1 < \cdots < V_n$, such that $V_i/V_{i-1} \cong A_i$ for $1 \le i < n$, and such that there exists no shorter subnormal series for $V$ in $W$. Write $A = A_0$ and let $G$ be the wreath product of $A$ with $W$ corresponding to the action of $W$ on the right cosets in $V$. In other words, $G = BW$ is a semidirect product, where $B \triangleleft G$ and $B$ is the direct product of $|W:V|$ copies of $A$. Also, $W$ acts to permute these direct factors of $B$, and this action is permutation isomorphic to the action of $W$ on the cosets of $V$ in $W$. (In fact, we assume that we are given a specific bijection from the set of cosets of $V$ onto the set of direct factors of $B$.)

Now let $C$ be the product of all of the direct factors of $B$ that correspond to nontrivial cosets of $V$, and note that ${\bf N}_W(C) = V$. Let $H = H_0$ be the group $CV$, and for $i > 0$, let $H_i = BV_i$. It is easy to see that $H_0 < H_1 < \cdots < H_n = G$ is a subnormal series with factors $A_i$ as wanted. We must show that no shorter subnormal series for $H$ exists. Note that the subnormal depth of $H_1$ is exactly $n - 1$. (This can be seen by intersecting a subnornal series for $H_1$ in $G$ with $W$. This yields a subnormal series for $V$ in $W$.)

Suppose $H \triangleleft K$. We argue that $BK = BV$. Otherwise, $BK > BV$, so $BK \cap W > V$. But $BK$ normalizes $C$ since $C = B \cap H$, and this contradicts the fact that $V$ is the full normalizer of $C$ in $W$. Now if $H = K_0 < K_1 < \cdots < K_m = G$ is a subnormal series for $H$, then $H_1 = BV = BK_1 \subseteq \cdots \subseteq BK_m = G$ is a subnormal series for $H_1$ with length at most $m-1$, and thus $m \ge n$, as wanted.


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