# Subgroup lattice

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.

• 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 – Jack Schmidt Feb 9 '14 at 19:51
• 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)? – Mike Miller Feb 9 '14 at 20:01
• The cyclic group of orders 2, 4, 8, and 16 are as you describe. The cyclic group of order 6 is a counterexample. – Jack Schmidt Feb 9 '14 at 20:08
• Okay thank you. I will check and see! – Mike Miller Feb 9 '14 at 20:14
• Ah yes, I see. I understand now - thanks for the help Jack. :) – Mike Miller Feb 9 '14 at 20:26

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.