Groups whose poset of direct factors are lattices

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.

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

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.

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:

(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.

(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.

(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 $$j$$.

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.

• Okay, I would just notice that a group satisfying both chain conditions on normal subgroups have unique Krull-Schmidt decomposition, up to the order of the factors, if and only if every direct factor have a unique normal complement. – Rajkarov Sep 16 '18 at 9:54
• @Rajkarov Well my initial expectation was that normal endomorphisms, and the $z$ morphism I use in particular, is the fundamental "thing" to consider here, and yields a rather concrete demonstration of what prevents the $\mathcal{D}$-group property from holding. So I went in that direction, ultimately to realize there's a few more fiddly bits with abelian direct factors to deal with (there always is, pretty much). I don't think it's actually difficult to detail this case, I just ran out of time and energy today to spend on it. – zibadawa timmy Sep 16 '18 at 10:10
• In the particular case of finite abelian groups, I think (if I'm not wrong) that a finite abelian group is a $\mathcal{D}$-group if and only if its sylow subgroups are indecomposable. – Rajkarov Sep 16 '18 at 10:20
• @Rajkarov Every subgroup of an elementary abelian $p$-group is a direct factor, so they're all $\mathcal{D}$-groups. It's the exponent, in the sense of wanting to (not) exhibit $z$ such that $\ker(z)$ is a proper non-trivial subgroup of an indecomposable abelian (thus cyclic of prime power order) group, which is relevant. – zibadawa timmy Sep 16 '18 at 10:32
• I agree with what you have said about the abelian case. – Keith Kearnes Sep 17 '18 at 12:55

I have a few observations about this question, but only time today to write down one of them. For this I will write $H\cap K$ for the intersection of two subgroups (just as everyone else does), and write $H+K$ for the join of the subgroups.

Theorem. Let $G$ be a group, and let $(P_1,Q_1)$ and $(P_2,Q_2)$ be two pairs of complementary normal subgroups (a.k.a. pairs of complementary direct factor subgroups of $G$). If $P = P_1\cap P_2$ and $Q = Q_1+Q_2$, then

1. $[G,G]\subseteq P+Q$.
2. $[P\cap Q,P+Q] = \{1\}$.

Therefore, if $G$ is any (finite) centerless, perfect group, then $G$ is a $\mathcal D$-group.

Proof: For the first item, $$\begin{array}{rl} [G,G]&=[P_1+Q_1,P_2+Q_2]\\ &=[P_1,P_2]+[P_1,Q_2]+[Q_1,P_2]+[Q_1,Q_2]\\ &\leq [P_1,P_2]+Q\\ &\leq (P_1\cap P_2) + Q = P+Q. \end{array}$$ Here I am using the additivity of the commutator, the fact that $[H,K]\leq H\cap K$, and the fact that $Q_1, Q_2\leq Q$.

For the second item, $[P,Q_1] \leq P\cap Q_1 \leq P_1\cap Q_1 = \{1\}$. Similarly $[P,Q_2] = \{1\}$. By the additivity of the commutator, $[P,Q]=[P,Q_1+Q_2]=[P,Q_1]+[P,Q_2]=\{1\}$. Now let $Z=P\cap Q$, which is $\leq P$ or $Q$. From the last two sentences and the monotonicity of the commutator in each variable we deduce $[Z,Q]\leq [P,Q] = \{1\}$ and $[Z,P]\leq [Q,P]=[P,Q]=\{1\}$, so by additivity we get $$[P\cap Q,P+Q]=[Z,P+Q]=[Z,P]+[Z,Q]=\{1\}.$$ This is the assertion to be proved.

For the final sentence of the proof, let $G$ be a perfect group ($[G,G]=G$) that is also a centerless group ($[G,N]=\{1\}$ implies $N=\{1\}$). Using the perfectness of $G$, the first item of the theorem can be written $G\subseteq P+Q$. Using this (i.e. $G=P+Q$), the second item can be written $[P\cap Q,G]=\{1\}$, or $P\cap Q\leq Z(G)$. Using the centerlessness of $G$ we get $P\cap Q=\{1\}$. Altogether we obtain that $P=P_1\cap P_2$ and $Q=Q_1+Q_2$ are complementary normal subgroups of $G$. This shows that the collection of factor congruences is closed under $\cap$ and $+$, so $G$ is a $\mathcal D$-group \\\

[One can go a bit further and show that the lattice of factor subgroups of a perfect, centerless group is a complemented distributive sublattice of ${\mathcal N}(G)$.]

I have a little remark. Every direct factor of $\mathcal{D}$-group is a $\mathcal{D}$-group. Indeed, let $H$ be a direct factor of a $\mathcal{D}$-group $G$. Clearly, a direct factor of $H$ is a direct factor of $G$ contained in $H$. Conversely, if $K$ is a direct factor of $G$ contained in $H$, then $G=K\times L$ for some subgroup $L$ of $G$. A small computation shows that $H=K\times(H\cap L)$ and then $K$ is a direct factor of $H$. We deduce then that the poset $\mathcal{D}(H)$ is isomorphic to the interval $[1,H]$ of $\mathcal{D}(H)$ which implies that $\mathcal{D}(H)$ is a sublattice of $\mathcal{N}(H)$.