# Given the lattice, find all pairs of elements that generate $D_{8}$

Given the subgroup lattice of the dihedral group $D_{8}$, find find all pairs of elements that generate $D_{8}$. There should be 8 pairs.

The given lattice is the image below:

I came to following set of pairs:

$\left \langle r, s \right \rangle$, $\left \langle r^2, s \right \rangle$, $\left \langle rs, r \right \rangle$, $\left \langle r^3s, r \right \rangle$, $\left \langle s, rs \right \rangle$, $\left \langle s, r^3s \right \rangle$, $\left \langle r^2s, rs \right \rangle$, $\left \langle r^2s, r^3s \right \rangle$

and

$\left \langle s, r^3 \right \rangle$, $\left \langle rs, r^3 \right \rangle$, $\left \langle r^3s, r^3 \right \rangle$, $\left \langle r^3, r^2s \right \rangle$

But I am not sure if this is right.

Also, how do you analyse the lattice step by step to reach the solution? Finding out first 8 was simple, because I had to find single-element subgroups and combine them to form pairs. However, the rest 4, I derived from found ones by searching for equivalent subgroups.

This is exercise 2.5.4 in Dummit & Foote "Abstract Algebra" (3rd ed.).

It's $12$, really.
For instance, if you take the pair of maximal subgroups $$\langle s, r^2 \rangle, \quad \langle r \rangle$$ you can choose either $s$ or $s r^2$ from the first subgroup, and either $r$ or $r^3$ from the second one, This makes for $4$ generating pairs.
As there are three pairs of maximal subgroups, and for distinct maximal subgroups $M, N$ you have $\lvert M \setminus M \cap N \rvert = 4 - 2 = 2$, you get $2 \cdot 2 \cdot 3 = 12$ (unordered) pairs.
• Why $r$ or $r^3$ from $<r>$? How can we tell from the lattice and not properties of $D_8$ directly that $r \in <r_3,s>$ so that it truly generated $D_8$? – topoquestion Sep 7 '16 at 14:49