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.
You have to choose first a pair of distinct maximal subgroups. And then one element from each of these maximal subgroup, neither of which elements lies in their intersection.
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.