Let $G$ be a finite group. Denote by $\mathcal{N}(G)$ the modular lattice of normal subgroups of $G$ and denote by $\mathcal{D}(G)$ the subposet of $\mathcal{N}(G)$ whose elements are the direct factors of $G$. 


In general, $\mathcal{D}(G)$ is not a sublattice of $\mathcal{N}(G)$. For example, take $G=\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ and $H,K$ the subgroups generated by $(1,0)$ and $(1,1)$ respectively. We can check that $H$ and $K$ are two direct factors of $G$ but that $H\cap K=<(2,0)>$ is not.


We will call $\mathcal{D}$-group every finite group $G$ for which $\mathcal{D}(G)$ is a sublattice of $\mathcal{N}(G)$. Some class of finite groups are $\mathcal{D}$-group. For example, we can check that :


1) Every cyclic group is a $\mathcal{D}$-group.


2) More generally, every finite nilpotent group for which the sylow subgroups are indecomposable is a $\mathcal{D}$-group.


3) Every group (finite or infinite) in which every normal subgroups is a direct factor is obviously a $\mathcal{D}$-group. It is well known that such groups are the restricted direct product of simple groups (see James. Weigold, On direct factors in groups).


My question is : Can we describe (or if it is possible, classify) the finite $\mathcal{D}$-group.


Remark : If the lattice of normal subgroups of a finite $\mathcal{D}$-group is distributive, then the lattice $\mathcal{D}(G)$ is a boolean algebra.



Groups whose poset of direct factors are lattices
*(edit: Added Theorem 2 below, which gives half the case with abelian direct factors, and classifies the finite abelian $\mathcal{D}$-groups)*



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> **Theorem 1**. Let $G$ be a non-trivial group satisfying both chain conditions on normal subgroups, and with no non-trivial abelian direct factors. Then $G$ is a $\mathcal{D}$-group if and only if $G$ admits a unique Krull-Schmidt decomposition, up to the order of the factors.
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*proof*: The chain conditions guarantee that $G$ admits a (finite length) Krull-Schmidt decomposition $G= G\_1\times \cdots \times G\_n$, where the subgroups $G\_1,...,G\_n$ are all non-trivial and indecomposable. Let $\pi\_1,...,\pi\_n$ denote the corresponding projections $\pi\_i\colon G\to G\_i$.


The group $\text{Aut}\_c(G) = C\_{\text{Aut}(G)}(\text{Inn}(G))$ acts transitively on the Krull-Schmidt decompositions, up to the order of the factors. $G$ admits a unique KS decomposition, up to order of the factors, if and only if $\text{Aut}\_c(G) = \prod\_{i=1}^n \text{Aut}\_c(G\_i)$. This is in turn equivalent to $\text{Hom}(G\_i,Z(G\_j))$ being trivial for all $i\neq j$. Which is in turn equivalent to $\pi\_j(\phi(G\_i))$ being trivial for all $i\neq j$ and $\phi\in\text{Aut}\_c(G)$.


So suppose that $G$ admits a unique KS decomposition, up to the order of the factors. Then every direct factor of $G$ is of the form $\prod\_{i\in E} G\_i$ for some $E\subseteq \{1,...,n\}$. The intersection and join of direct factors are therefore equivalent to the intersection and union of the corresponding subsets of $\{1,...,n\}$. Therefore the direct factors form a sublattice of $\mathcal{N}(G)$, and $G$ is a $\mathcal{D}$-group.


On the other hand, suppose that $G$ does not admit a unique KS decomposition up to the order of the factors. Fix any KS decomposition $G= G\_1\times\cdots G\_n$. Also fix $i,j$ such that $\text{Hom}(G\_i,Z(G\_j))$ is non-trivial, and let $z\in\text{Hom}(G\_i,Z(G\_j))$ be non-trivial. We define $\phi\in\text{Aut}\_c(G)$ by $\phi(g)=g$ for $g\in G\_k\neq G\_i$ and $\phi(g)= g z(g)$ for all $g\in G\_i$. Then $G\_i$ and $\phi(G\_i)$ are distinct direct factors of $G$ but $G\_i\cap \phi(G\_i) =\ker(z)$ is a proper normal subgroup of the indecomposable group $G\_i$, so can be a direct factor only if $\ker(z)=1$. This implies $G\_i$ is abelian, a contradiction to assumptions on $G$. $\square$


The reverse direction did not use the assumption of no abelian direct factors.



> 
> **Theorem 2**. Let $G$ be a non-trivial group satisfying both chain conditions on normal subgroups. Write $G= (A\_1\times\cdots\times A\_k)\times (G\_1\times \cdots \times G\_n)$, and $A=A\_1\times\cdots \times A\_k$, where the $A\_i$ are indecomposable abelian groups and the $G\_i$ are indecomposable non-abelian groups. If $G$ is a $\mathcal{D}$-group then the following three conditions hold:
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> (1) The Sylow subgroups of $A$ are either cyclic or elementary abelian. Equivalently, for all $i\neq j$, any non-trivial element of $\text{Hom}(A\_i,A\_j)$ is an injection.
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> (2) The Krull-Schmidt decomposition of $G\_1\times\cdots\times G\_n$ is unique, up to the order of the factors. Equivalently, for all $i\neq j$, $\text{Hom}(G\_i,Z(G\_j))$ is trivial.
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> (3) For all $i,j$, any non-trivial element of $\text{Hom}(A\_i,Z(G\_j))$ is an injection. Given the previous two conditions, this is equivalent to saying that if the Sylow $p$-subgroup of $A$ is not elementary abelian, then the Sylow $p$-subgroup of $Z(G\_j)$ is trivial for all $j$.
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*proof*: The proof proceeds essentially as in proof of the forward direction for the previous proof.


We write $G=H\_1\times\cdots \times H\_m$, where we permit one or more factors to be abelian. For any $i\neq j$ we consider $z\in\text{Hom}(H\_i,Z(H\_j))$, and define $\phi\in\text{Aut}\_c(G)$ by $\phi(g)=g$ for all $g\in H\_k\neq H\_i$ and $\phi(g)=g z(g)$ for all $g\in H\_i$. We consider $H\_i\cap \phi(H\_i)=\ker(z)$, which is the intersection of two direct factors of $G$.


If $G$ is a $\mathcal{D}$-group, then $\ker(z)$ is a normal subgroup of the indecomposable group $H\_i$ which is also a direct factor of $G$. Therefore either $\ker(z)=H\_i$ or $\ker(z)=1$. In particular, for all $i\neq j$ any non-trivial element of $\text{Hom}(H\_i,Z(H\_j))$ must be an injection. If such an injection exists, then $H\_i$ must be abelian. As every quotient of a finite abelian group $X$ is isomorphic to a subgroup of $X$, and vice versa, the three conditions then follow. $\square$.


I'm fairly confident the converse also holds, thereby giving the full classification of $\mathcal{D}$-groups with both chain conditions, and so in particular all finite $\mathcal{D}$-groups. But I'm still working on that.


