# 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 $\mathcal{N}(H)$.