Visual example on Centralizer of $D_4, D_3$? In a text, given definition of Centralizer in context of $D_4$ as: If $x \in D_4$, the centralizer of $x$, denoted by $CD_4(x)$, is the set of all elements in $D_4$ that commute with $x$.
 First of all, I am not clear with the defn. as no example (visual in particular) is given. So, if some examples are referenced or given, then can understand it.
 Second, there is a small exercise immediately that states: 
Let $D_3$ denote the set of symmetries of an equilateral triangle. Find the
 multiplication table for $D_3$. What is the center of $D_3$?
Addendum: The answer is given for $D_4, D_3$ center as $ = \{R_0, R_{180}\}$, and respectively $\{e\}$, where $e$ is the identity element, i.e. the start position or null rotation.
I can easily find the multiplication table for $D_3$ around the medians, given the 3 symmetries and 3 possible rotations & reflections. Have a small doubt as to why there is no identity action in terms of reflection in $D_3$. I mean that the reason could gather is that, the reflections are not transforming from one to the other (reflection), and need rotations only to move from one to another, In other words, the 3 reflections are not capable of transforming by just reflection into each other, & hence back to the original one. Further all 3 are equivalent as being of the same type, if reflection is unable to go from one to any other, hence differing from rotations where there is one starting or identuty position. further can be described in terms of still basic action of rotation.
 A: In the comments, OP has restated the question as, "What should be a suitable way to view / know sets as center, centralizer (given the symmetry properties, and the set of elements in an object's group), with out having to explicitly draw the group table?" 
Let's look at $D_4$. Let $r$ be a rotation one-fourth of the way around. Any element of any group commutes with all of its powers, so the centralizer of $r$ contains $H=\{\,1,r,r^2,r^3\,\}$. I want to show the centralizer is exactly $H$. Note that the centralizer is a subgroup of $D_4$, and the order of a subgroup divides the order of the group, so if the centralizer isn't $H$, it must be all of $D_4$. So all we have to do is find one element of $D_4$ that doesn't commute with $r$, and then we can conclude the centralizer is $H$. 
Now it's easy enough to take any element $s$ that's not in $H$ and by computation show $rs\ne sr$. But you want to do it without using the group table. I'm not sure what you do want to allow, and I'm not sure I can do it without calculating $rs$ and $sr$ and seeing that they are different. 
Once you know the centralizer of $r$ is $H$, let $s$ be any element not in $H$; it doesn't commute with $r$, nor with $r^3=r^{-1}$, but it does commute with $r^2$, so the center is $Z=\{\,1,r^2\,\}$. Again, I don't know how to do this without using the group table to do some calculations. 
