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.
We can assume $n > 1$, and we induct on $n$. By the inductive hypothesis, let $U$ 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. 1 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$U$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.