lattice of subgroup

Question

In the spirit of category theory, we use relations between objects to describe the object themselves. In the same sense I want to use the lattice structure of subgroups to decide whether a subgroup is normal. We know that if $H \lhd G$ and $K \subset G$, then there is a lattice preserving bijection from subgroups of $K$ containing $H \cap K$ to subgroups of $HK$ containing $H$. This gives us a categorical description of normality.

My question is: Does the converse hold? That is, if for every $K \subset G$, this bijection also holds, must $H$ be normal in $G$? (Here we need a modification: $HK$ needs to be replaced by $\langle H, K \rangle$.)

Hmm, so certainly $HK$ being a subgroup for all $K$ puts some restrictions on $H$ to begin with. I need to think a bit more about how much. — Tobias Kildetoft, Jan 30 '17 at 10:04
@TobiasKildetoft Here $HK$ denotes the minimum subgroup which contains $H$ and $K$. — Minghui Ouyang, Jan 30 '17 at 10:11
Please don't write $HK$ then. Write $\langle H,K\rangle$. The former is not the subgroup generated by them. — Tobias Kildetoft, Jan 30 '17 at 10:18
Actually, as in lattices in general, it is common to denote the supremum of $H$ and $K$ by $H\vee K$. — Stefan Perko, Jan 30 '17 at 15:26
Possible duplicate of What Do Normal Subgroups Look Like in the Lattice of Subgroups? — verret, Jan 30 '17 at 19:19

score 6 · Accepted Answer · answered Jan 30 '17 at 18:07

It's impossible to identify the normal subgroups of $G$ just by looking at the lattice of subgroups of $G$.

For an example, let's take the simplest non-abelian group, $S_3$. Its subgroup lattice consists of $S_3$ at the top, the trivial group $\{e\}$ at the bottom, and four intermediate groups: $A_3$, which is generated by the permutation $(123)$, and the $2$-element subgroups generated by $(12)$, $(13)$, and $(23)$, respectively. Any two intermediate subgroups have join $S_3$ and meet $\{e\}$, so the map which swaps any two of them is a lattice isomorphism. That is, they are indistinguishable as elements of the lattice. But $A_3$ is normal, while the other three are not.

William DeMeo · Answer 2 · 2017-02-06T04:20:45.353

Even after reading Alex Kruckman's nice answer, the OP might still wonder if the structure of a subgroup lattice could give some information about the location, or number, of normal subgroups. Generally speaking, it cannot.

For example, there is an abelian group with the very same subgroup lattice as the one in Alex's example---namely, the group $\mathbb Z/3 \times \mathbb Z/3$. Of course, all subgroups are normal in this case (unlike in Alex's example).

On the other hand, there are certainly special cases in which the subgroup lattice tells us which subgroups are normal, but usually that's a consequence of the lattice structure telling us much more. For example, the group $A_4$ can be uniquely identified by its subgroup lattice. That is, no other group has the same subgroup lattice. And in that case, we know that the only nontrivial proper normal subgroup happens to be the top of the $M_3$ sublattice of the subgroup lattice of $A_4$ (but this was "a priori" information---we didn't derive it directly from the lattice structure per se).

As another trivial example, when a subgroup lattice is a chain, every subgroup is normal. In this case, we can deduce this information from the shape of the subgroup lattice. (If any of the subgroups were non-normal, they would have conjugate subgroups at the same height in the lattice.)

There are many more examples like these, and much more to say about what properties of a group can be inferred from the structure of its subgroup lattice. See Roland Schmidt's book "Subgroup Lattices of Groups."

lattice of subgroup

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