Finding all minimal generating sets of $D_4$. From VGT By Nathan Carter 
I have computed a), b), c) parts of question above. I am stuck on the d) part.   

Now clearly given in the question are two minimal sets having $2$ elements each, but how do I find out other $10$. Do I have to use Cayley diagram for that? If yes, then how?

My $D_4:=\{f,e,r,r^2,r^3,r^3f,r^2 f, rf\}$.
Thanks
 A: I am going to give you the answer for $D_6$ instead of $D_4$ so you can fill the details. This method just relies on the definitions of $D_n$.
Any group with 1 generator is abelian. In $D_6$ you have that for any rotation $r$ and a reflection $s$, $(rs)^2 = 1$, since a rotation times a reflection is a reflection since it has a fixed point or a fixed edges (when the action can be also be understood is in the set of edges of the n-gon). Now $D_6$ can't be abelian since this would mean that $1 = (rs)^2 = r^2$ for any rotation, but there is at least one rotation of order $6$. Then a minimal set would have at least 2 generators. 
Consider $r,s$ where $r,s$ are as above and $r$ has order 6 but not 2 or 3. Note that the identity $(rs)^2 = 1$ gives us a way of expressing any product of $s$'s and $r$'s in the form $s^i r^j$. Let
$$G = \{ 1, r, r^2,\dots, r^5, s, sr, \dots, sr^5\}$$
Contains at most 12 elements, if $s^ir^j = s^l r^k$ for some elements in $G$, in the cases where $i = l$ we have $r^j = r^k$ which means $j = k$. If $sr^k = r^j$ then $s r^n = 1$ for some $n$, this would mean that the rotation $r^{-n}$ is a reflection, but this cannot be as reflections have fixed points (or edges) and rotations don't, so $G=D_6$. 
Any set of the form $\{r,s\}$ as above is a minimal set of generators and generating set must have a reflection, as the product of rotations only gives rotations. Since there are  exactly two rotations $r,r^5$ of order 6 and there are 6 reflections $sr^i$, We are left to count the cases $\{s,r\}$ where $r$ is a rotation of order 2 or 3, there cant be any such set with $r$ of order 2 as $D_6$ has elements of order 3, and when the order is 2, one can see that we can only obtain rotations of order 2 so thats not possible either, i.e. there are 12 possible sets with one rotation and one reflection. The only things left to count is pairs of reflections, let us start with the ones containing a fixed reflection $s$, we have as our only options $\{s,sr\}, \{s,sr^5\}$ are the only possibilities, changing the choice of $s$ we have a total of 12 options, where we have to remove duplicates, i.e. 6 options since each option repeats once.
