TITLE: Must the centralizer of an element of a group be abelian?
QUESTION [16 upvotes]: Must the centralizer of an element of a group be abelian? 
I see that the definition of centralizer is: 

Let $a$ be a fixed element of a group $G$. The centralizer of $a$ in $G$, $C(a)$, is the set of all elements in $G$ that will commute with $a$. In symbols, $C(a)=\{g\in G \mid ga=ag\}$.

But this doesn't necessarily mean that the centralizer is abelian, does it?

REPLY [5 votes]: An example that is a specific instance of Orat's answer, in the setting of plane geometry.
Take the dihedral group $D_{2n}$ of symmetries of a regular polygon with an even number of sides: it consists of rotations  of the plane by angles that are multiples of $\pi/n= \frac{2\pi}{2n}$, and $2n$ reflections ($n$ of them  about  diagonals connecting opposite pair of vertices, and $n$ more about lines connecting mid-points of pairs of opposite sides). 
You can check that this group is non-abelian. But the rotation by the specific angle $\pi$, has the whole group as its centralizer.
Easy to verify bu taking $n=4$, corresponding to the symmetries of the square shape.