Generalisation of notion of maximal subgroup / maximal subalgebra

Question

I'm trying to trace a definition. For this question fix some group $G$ (we generalise to an arbitrary algebra later).

Definitions:

1. Write $G'\leq G$ when $G'$ is a subgroup of $G$.
2. Call $G'$ a maximal proper subgroup when $G'\leq G$ and $G'\neq G$ and $G'$ is maximal such.
3. If $X\subseteq G$ then write $\langle X\rangle$ for the least subgroup of $G$ containing $X$, and call this the subgroup generated by $X$.
4. If $X,Y\subseteq G$ then write $X\bowtie Y$ when $\langle X\cup Y\rangle=G$. (So $X$ generates $G$ when $X\bowtie X$.)

We now come to something less familiar:

Definition: Call a subgroup $T\leq G$ cotopen when for every pair of subgroups $G',G''\leq G$ we have that $$ G'\bowtie G'' \quad\text{implies}\quad (G'\bowtie T) \vee (T\bowtie G'') . $$ Unpacking the notation, this means that if $G'\cup G''$ generates $G$ then so does $G'\cup T$ or $T\cup G''$.

The property of being cotopen can be viewed as generalising the notion of maximal subgroup, in the following sense:

Lemma: If $T\leq G$ is a maximal proper subgroup of $G$, then $T$ is cotopen.

Proof: Consider any $G',G''\leq G$ such that $G'\bowtie G''$. There are now two easy cases:

• If $G'\subseteq T$ then from $G'\bowtie G''$ it easily follows that $T\bowtie G''$ and we are done.
• If $G'\not\subseteq T$ then by maximality of $T$ we have that $G'\bowtie T$ and again we are done.

My questions are:

1. Has cotopenness been studied already and if so, where?
2. Does this seem an interesting thing to look at?

Note on generalisations: I set this question up for groups but clearly the definitions work for subalgebras of any algebraic structure, and I am equally interested in cotopenness for arbitrary algebras. Cotopenness almost certainly generalises further (though I have not yet checked it) --- e.g. to the (maximal) ideals of a ring --- and I am also interested in that.

Thank you.

Answer 1

I'm elaborating on Arturo Magidin's excellent answer. Fix a group $G$.

Definitions:

1. If $G',G''\leq G$ then call $G'$ an overgroup of $G''$ when $G''\leq G' \leq G$. Call an overgroup $G'$ proper when $G''\leq G'\neq G$.
2. Write $POver(G'')$ for the poset of proper overgroups of $G''$ in $G$, ordered by $\leq$. Note that $POver(G'')$ is just the lattice of overgroups of $G''$ (with join $\langle -,-\rangle$), but with the top element $G$ removed.
3. If $G''\leq G$ then call $G'\in POver(G'')$ maximal when $G'$ is a maximal element in $POver(G'')$ (so it is a maximal proper subgroup of $G$ that contains $G''$). Similarly, call $G'$ greatest when it is a greatest element (= maximal and unique such).
4. Call $POver(G'')$ directed when if $G_1,G_2\in POver(G'')$ then $\langle G_1,G_2\rangle\in POver(G'')$.

Claim: $T$ is cotopoen if and only if $POver(T)$ is directed.

Proof. We prove two contrapositive implications:

• Suppose there exist overgroups $E_1,E_2\in POver(T)$ of $T$ such that $\langle E_1, E_2\rangle=G$. Clearly $\langle T,E_2\rangle = E_2$ and $\langle E_1,T\rangle = E_1$. Thus, $T$ is not cotopen.

• Suppose $T$ is not cotopoen. Let $H$ and $K$ be subgroups such that $\langle H,K\rangle = G$, but $E_1=\langle H,T\rangle \neq G$ and $E_2=\langle K,T\rangle\neq G$. Then we see that $POver(T)$ is not directed, because it contains $E_1$ and $E_2$ but not $\langle E_1,E_2\rangle\geq \langle H,K\rangle=G$.

Corollary of Claim: If $G$ satisfies ACC on subgroups (Ascending Chain Condition: every ascending chain is eventually constant) then $T$ is cotopoen if and only if $T$ has a greatest (= unique maximal) proper overgroup. In particular this holds if $G$ is finite.

Proof: It is a fact that a directed poset with ACC has a greatest element.

Remark: For finite groups, "cotopen" = "has unique maximal proper overgroup".

Question: Assuming the analysis above is correct, I can repose my original question as follows: is the study of (unique, maximal) proper subalgebras above some subalgebra (or perhaps of similar structures, like ideals) a thing, and if so, can someone provide references? Thank you.

Answer 2

Too long for a comment.

Note that if $T$ is cotopoen then so is every subgroup that contains $T$.

Claim If $G$ is finite and has at least two maximal subgroups, then a subgroup $T$ is cotopen if and only if it is not contained in the intersection of any two maximal subgroups of $G$.

Proof. If there exist maximal subgroup $M_1$ and $M_2$ of $G$ with $M_1\neq M_2$ and $T\leq M_1\cap M_2$,, then $G=\langle M_1,M_2\rangle$, but $\langle T,M_2\rangle = M_2$ and $\langle M_1,T\rangle = M_1$. So $T$ is not cotopen.

Conversely, if $T$ is not cotopoen, then let $H$ and $K$ be subgroups such that $\langle H,K\rangle = G$, but $\langle H,T\rangle \neq G$, $\langle K,T\rangle\neq G$. Let $M_1$ and $M_2$ be maximal subgroups such that $\langle H,T\rangle \leq M_1$ and $\langle K,T\rangle\leq M_2$. Note that since $\langle H,K\rangle = G$, then $M_1\neq M_2$; and $T\leq M_1\cap M_2$, as desired. $\Box$

The same holds if $G$ satisfies ACC on subgroups.