# 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.

• Is my question too simple? – jiten Jan 6 '18 at 10:53
• It's not clear to me whether you understand that $D_3$ has six elements, the three rotations (counting the identity element as a rotation) and the three reflections. It's also not clear to me whether you know what is meant by the center of a group. Now, if you know the eight elements of $D_4$, then for each pair of elements you can calculate $xy$ and $yx$; if they are the same, then $y$ is in the centralizer of $x$ – otherwise, not. – Gerry Myerson Jan 8 '18 at 2:54
• If by "visual" you mean drawings/diagrams, I'm no good at posting those here. Anyway, suppose $x$ in $D_4$ is rotation one-fourth of the way around. Then all the rotations commute with it (that is, if $y$ is a rotation, then $xy=yx$, but the reflections don't commute with it (that is, if $y$ is not a rotation, then $xy\ne yx$, so the centralizer of $x$ is the set of four rotations. In $D_3$, since the non-identity rotations don't commute with any of the reflections, the only element in the center is the identity. – Gerry Myerson Jan 8 '18 at 21:20
• I've written a partial answer to your restated question. But I don't know what you mean by "a group's elements are mapped in a composition of two (set) elements (actions) under a (composition) operator such that their tabular representation is a Latin square." Maybe you could show me what this square looks like (it sounds like the group table to me). Just knowing the group table is a Latin square doesn't get you very far, as there are lots of $8\times8$ Latin squares, and most of them don't correspond to any group, and some of them correspond to abelian groups and not to $D_4$. – Gerry Myerson Jan 11 '18 at 9:28
• I also don't understand "a theorem that is based on the occurrence of unique solution in linear algebra (matrix) form ($AX=B$) of it" It sounds like you are talking about group representations, but doing it out in detail would help me see what you mean. – Gerry Myerson Jan 11 '18 at 9:30

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.