# Classification of indecomposable inclusions $(H \subset G)$ with $G$ decomposable

Definition: A group $G$ is indecomposable if: $G = G_1 \times G_2 \Rightarrow \exists i \ G_i = 1$.

We can generalize the notion of indecomposable from groups to inclusion of groups as follows:

Definition: Two inclusions of finite groups are equivalent, $(A \subset B) \sim (C \subset D)$, if: $(A/A_B \subset B/A_B) \simeq (C/C_D \subset D/C_D)$ with $A_B$ the normal core of $A$ in $B$.
Remark: The equivalence class of $(A \subset B)$ is the same that the conjugacy class of a transitive permutation group $G$ (see this GAP Data Library) with $(A \subset B) \sim (G_1 \subset G)$.

Definition: An inclusion of groups $(H \subset G)$ is indecomposable if (for $H_i \le G_i)$: $(H \subset G) \sim (H_1 \times H_2 \subset G_1 \times G_2)$ $\Rightarrow$ $\exists i \ H_i = G_i$
Examples: The maximal inclusions are indecomposable.

In the maximal (finite index) case, $(H \subset G)$ is indecomposable and $G$ decomposable if and only if
$(H \subset G) \sim (D_S \subset S \times S)$ with $S$ a nonabelian finite simple group and $D_S$ the diagonal subgroup.
It's proved by the answer here. Are there others examples beyond the maximal case?

Question: What's the classification of indecomposable inclusions $(H \subset G)$ with $G$ decomposable?

If I have understood the definition of an indecomposable inclusion correctly, then any core-free subgroup $H$ of a direct product $G=G_1 \times G_2$ in which $H$ is indecomposable, and is not contained in any direct factor of $G$ isomorphic to $G_1$ and $G_2$ will be an example.
There are many such examples, such as $G=S_3 \times S_3$ with $H$ a diagonal subgroup of order $2$, $3$, or $6$.
• It seems to be essentially a diagonal phenomenon. If $G_1$ and $G_2$ have no isomorphic (non-trivial) subgroup then an indecomposable subgroup $H$ of $G_1 \times G_2$ is contained in one direct factor. Then we could reduce the classification to the classification of $(H \subset K^n)$ with $H$ core-free and $H \le D_K$ with $D_K$ the whole diagonal subgroup of $K^n$, isn't it? – Sebastien Palcoux Aug 31 '14 at 15:12
• About the choice of this long definition, what is not clear to me is the existence of a core-free decomposable inclusion $(H \subset G)$ which is (then) $\sim (H_1 \times H_2 \subset G_1 \times G_2)$ but $\not = (H'_1 \times H'_2 \subset G'_1 \times G'_2)$ (with $H'_i < G'_i$). – Sebastien Palcoux Aug 31 '14 at 15:30
• About the first comment: "$G_1$ and $G_2$ have no isomorphic (non-trivial) subgroup" means that $1 \neq K_i \le G_i$ implies that $K_1 \not\simeq K_2$. – Sebastien Palcoux Aug 31 '14 at 15:34
• The first sentence of your first comment is correct if $G_1$ and $G_2$ are both finite, because then they must have coprime order, but not in general. For example, the (additive) group ${\mathbb Z}_4 \oplus {\mathbb Z}$ has infinite cyclic subgroups that are not contained in either factor. – Derek Holt Aug 31 '14 at 17:42