# lattice of subgroups

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 $\langle X\rangle$, 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 $\langle e\rangle$ 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

 $\bigwedge_{i\in I}H_{i}\qquad\mbox{ and }\qquad\bigvee_{i\in I}H_{i}$

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.

Example. Note that $\operatorname{Aut}\mathbb{Z}_{p^{2}}\cong\mathbb{Z}_{p-1}\times\mathbb{Z}_{p}$. Therefore it is possible to from a non-trivial semidirect product  $\mathbb{Z}_{p^{2}}\rtimes\mathbb{Z}_{p}$. The lattice of subgroups of $\mathbb{Z}_{p^{2}}\rtimes\mathbb{Z}_{p}$ is the same as the lattice of subgroups of $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. However, $\mathbb{Z}_{p^{2}}\rtimes\mathbb{Z}_{p}$ is non-abelian   while $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ is abelian  so the two groups are not isomorphic.

Similarly, the groups $\mathbb{Z}_{p^{i}}\rtimes\mathbb{Z}_{p}$ 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, 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 $p|q-1$, $\mathbb{Z}_{p}\leq\mathbb{Z}_{q}^{\times}$ so if we write $\mathbb{Z}_{p}=\langle\omega\rangle$ 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 $0\leq i\leq p-1$ we get an automorphism $f_{i}:\mathbb{Z}_{q}\rightarrow\mathbb{Z}_{q}$ given by

 $f_{i}(n)=\omega^{i}n.$

Therefore to every $0\leq i\leq p-1$ we can define a group $G_{i}=\langle V,g_{i}\rangle$, $g_{i}=f_{1}\oplus f_{i}$ as a subgroup of $AGL(V)$. That is to say, $G_{i}=V\rtimes\langle g_{i}\rangle$ where the action of $g_{i}$ on $V$ is given by: for every $v\in V$, $v=\begin{bmatrix}n\\ m\end{bmatrix}$ for $n,m\in\mathbb{Z}_{q}$ set

 $g_{i}(v)=\begin{bmatrix}\omega&0\\ 0&\omega^{i}\end{bmatrix}\begin{bmatrix}n\\ m\end{bmatrix}=\begin{bmatrix}\omega n\\ \omega^{i}m\end{bmatrix}.$

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 LatticeOfSubgroups 2013-03-22 15:47:42 2013-03-22 15:47:42 CWoo (3771) CWoo (3771) 14 CWoo (3771) Definition msc 20E15 subgroup lattice