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](http://groupprops.subwiki.org/wiki/Normality_is_not_transitive_for_any_pair_of_nontrivial_quotient_groups). 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](http://groupprops.subwiki.org/wiki/Base_of_a_wreath_product_implies_right-transitively_2-subnormal), for instance.



Finding a subnormal series with specified quotients and end group of specific depth (defect)
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.


