lattice of subgroups
Let $G$ be a group and $L(G)$ be the set of all subgroups^{} of $G$. Elements of $L(G)$ can be ordered by the set inclusion relation^{} $\subseteq $. This way $L(G)$ becomes a partially ordered set^{}.
For any $H,K\in L(G)$, define $H\wedge K$ by $H\cap K$. Then $H\wedge K$ is a subgroup of $G$ and hence an element of $L(G)$. It is not hard to see that $H\wedge K$ is the largest subgroup of both $H$ and $K$.
Next, let $X=H\cup K$ and define $H\vee K$ by $\u27e8X\u27e9$, the subgroup of $G$ generated by $X$. So $H\vee K\in L(G)$. Each element in $H\vee K$ is a finite product^{} of elements from $H$ and $K$. Again, it is easy to see that $H\vee K$ is the smallest subgroup of $G$ that has $H$ and $K$ as its subgroups.
With the two binary operations^{} $\wedge $ and $\vee $, $L(G)$ becomes a lattice^{}. It is a bounded lattice^{}, with $G$ as the top element and $\u27e8e\u27e9$ as the bottom element. Furthermore, if $\{{H}_{i}\mid i\in I\}$ is a set of subgroups of $G$ indexed by some set $I$, then both
$$\underset{i\in I}{\bigwedge}{H}_{i}\mathit{\hspace{1em}\hspace{1em}}\text{and}\mathit{\hspace{1em}\hspace{1em}}\underset{i\in I}{\bigvee}{H}_{i}$$ |
are subgroups of $G$. So $L(G)$ is a complete lattice^{}. From this, it is easy to produce a lattice which is not a subgroup lattice of any group.
Atoms in $L(G)$, if they exist, are finite cyclic groups^{} of prime order (or $\mathbb{Z}/p\mathbb{Z}$, where $p$ is a prime), since they have no non-trivial proper subgroups^{}.
Remark. Finding lattices of subgroups of groups is one way to classify groups. One of the main results in this branch of group theory states that the lattice of subgroups of a group $G$ is distributive (http://planetmath.org/DistributiveLattice) iff $G$ is locally cyclic.
It is generally not true that the lattice of subgroups of a group determines the group up to isomorphism^{}. Already for groups of order ${p}^{3}$, $p>2$ or ${p}^{4}$, for all primes, there are examples of groups with isomorphic (http://planetmath.org/LatticeIsomorphism) subgroup lattices which are not isomorphic groups^{}.
Example. Note that $\mathrm{Aut}{\mathbb{Z}}_{{p}^{2}}\cong {\mathbb{Z}}_{p-1}\times {\mathbb{Z}}_{p}$. Therefore it is possible to from a non-trivial semidirect product^{} ${\mathbb{Z}}_{{p}^{2}}\u22ca{\mathbb{Z}}_{p}$. The lattice of subgroups of ${\mathbb{Z}}_{{p}^{2}}\u22ca{\mathbb{Z}}_{p}$ is the same as the lattice of subgroups of ${\mathbb{Z}}_{{p}^{2}}\times {\mathbb{Z}}_{p}$. However, ${\mathbb{Z}}_{{p}^{2}}\u22ca{\mathbb{Z}}_{p}$ is non-abelian^{} while ${\mathbb{Z}}_{{p}^{2}}\times {\mathbb{Z}}_{p}$ is abelian^{} so the two groups are not isomorphic.
Similarly, the groups ${\mathbb{Z}}_{{p}^{i}}\u22ca{\mathbb{Z}}_{p}$ and ${\mathbb{Z}}_{{p}^{i}}\times {\mathbb{Z}}_{p}$ for any $i>2$ and any primes $p$ also have isomorphic subgroup lattices while one is non-abelian and the other abelian. So this is indeed a family of counterexamples.
Upon inspecting these example it becomes clear that the non-abelian groups^{} have a different sublattice of normal subgroups^{}. So the question can be asked whether two groups with isomorphic subgroup lattices including matching up conjugacy classes^{} (so even stronger than matching normal subgroups) can be non-isomorphic groups. Surprisingly the answer is yes and was the dissertation of Ada Rottländer[1], a student of Schur’s, in 1927. Her example uses groups already discovered by Otto Hölder in his famous classification of the groups of order ${p}^{3}$, ${p}^{2}q$, and ${p}^{4}$. With the modern understanding of groups the counterexample is rather simple to describe – though a proof remains a little tedious.
Let $V={\mathbb{Z}}_{q}^{2}$ where $q$ is a prime – that is $V$ is the 2-dimensional vector space^{} over the field ${\mathbb{Z}}_{q}$. Let $p|q-1$ be another prime. As $p|q-1$, ${\mathbb{Z}}_{p}\le {\mathbb{Z}}_{q}^{\times}$ so if we write ${\mathbb{Z}}_{p}=\u27e8\omega \u27e9$ multiplicatively so that we have for every $n\in {\mathbb{Z}}_{q}$, $n\mapsto \omega \cdot n$ is an automorphism^{} of ${\mathbb{Z}}_{q}$. (Note that $\omega $ is often called a primitive $p$-th root of unity in ${\mathbb{Z}}_{q}$ as it spans the ${\mathbb{Z}}_{p}$ subgroup of ${\mathbb{Z}}_{q}^{\times}$.) Furthermore, for any $0\le i\le p-1$ we get an automorphism ${f}_{i}:{\mathbb{Z}}_{q}\to {\mathbb{Z}}_{q}$ given by
$${f}_{i}(n)={\omega}^{i}n.$$ |
Therefore to every $0\le i\le p-1$ we can define a group ${G}_{i}=\u27e8V,{g}_{i}\u27e9$, ${g}_{i}={f}_{1}\oplus {f}_{i}$ as a subgroup of $AGL(V)$. That is to say, ${G}_{i}=V\u22ca\u27e8{g}_{i}\u27e9$ where the action of ${g}_{i}$ on $V$ is given by: for every $v\in V$, $v=\left[\begin{array}{c}\hfill n\hfill \\ \hfill m\hfill \end{array}\right]$ for $n,m\in {\mathbb{Z}}_{q}$ set
$${g}_{i}(v)=\left[\begin{array}{cc}\hfill \omega \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {\omega}^{i}\hfill \end{array}\right]\left[\begin{array}{c}\hfill n\hfill \\ \hfill m\hfill \end{array}\right]=\left[\begin{array}{c}\hfill \omega n\hfill \\ \hfill {\omega}^{i}m\hfill \end{array}\right].$$ |
We are now prepared to give the Rottländer counterexample.
Now let $a$ and $b$ be integers between $2$ and $p-1$ such that $a$ is not congruent to $b$ modulo $p$. Notice this already forces $p>3$ so our smallest example is $q=11$ and $p=5$. Then ${G}_{a}$ is not isomorphic to ${G}_{b}$ (compare the eigenvalues^{} of ${g}_{a}$ to ${g}_{b}$ – they are not equal so the linear transformations are not conjugate in $GL(2,q)$.) However, ${G}_{a}$ and ${G}_{b}$ have isomorphic subgroup lattices including matching conjugacy classes.
References
- 1 Rottländer, Ada, Nachweis der Existenz nicht-isomorpher Gruppen von gleicher Situation der Untergruppen, Math. Z. vol. 28, 1928, 1, pp. 641– 653, ISSN 0025-5874. MR MR1544982,
Title | lattice of subgroups |
---|---|
Canonical name | LatticeOfSubgroups |
Date of creation | 2013-03-22 15:47:42 |
Last modified on | 2013-03-22 15:47:42 |
Owner | CWoo (3771) |
Last modified by | CWoo (3771) |
Numerical id | 14 |
Author | CWoo (3771) |
Entry type | Definition |
Classification | msc 20E15 |
Synonym | subgroup lattice |