1
$\begingroup$

I've been searching around for a while now and can't seem to find a clear explanation of what a subgroup lattice of a group actually is. I see the vertex set is given by the subgroups of the group, but how is the edge set related? My guess is that vertices are connected if one is a subset of the other - for example a subgroup containing just the identity would be connected to everything.

$\endgroup$
5
  • 2
    $\begingroup$ The edge relation is called "covering", there is an edge from $H$ to $K$ iff $H < K$ but there is no $L$ with $H<L<K$. en.wikipedia.org/wiki/Hasse_diagram $\endgroup$ Feb 9, 2014 at 19:51
  • $\begingroup$ I think I see. So say if you had subgroups of all different orders, it would just be a straight line (as they would all contain the identity)? $\endgroup$ Feb 9, 2014 at 20:01
  • 1
    $\begingroup$ The cyclic group of orders 2, 4, 8, and 16 are as you describe. The cyclic group of order 6 is a counterexample. $\endgroup$ Feb 9, 2014 at 20:08
  • $\begingroup$ Okay thank you. I will check and see! $\endgroup$ Feb 9, 2014 at 20:14
  • $\begingroup$ Ah yes, I see. I understand now - thanks for the help Jack. :) $\endgroup$ Feb 9, 2014 at 20:26

1 Answer 1

2
$\begingroup$

A lattice is a partially ordered set in which every two element has a least upper and a greatest lower bound (these are the lattice operations).

The elements of the subgroup lattice of a group are indeed the subgroups, and the partial order is defined by containment, so if you want to depict it you can draw an arrow from any subgroup $H_1$ to any other subgroup $H_2$ that contains it.. (If $G$ is finite, it is enough to draw down the coverings as @JackSchmidt commented, as, by transitivity it will generate the whole lattice.)

Anyway, the greatest lower bound of subgroups $H_1,\,H_2$ is $\ H_1\cap H_2$, and their least upper bound is the subgroup $\langle H_1\cup H_2\rangle$ generated by their union.

$\endgroup$
1
  • $\begingroup$ Thank you, the explanation helped. :) $\endgroup$ Feb 9, 2014 at 20:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .