Must the centralizer of an element of a group be abelian? 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?
 A: 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.
A: Let $G$ be any non-abelian group, and let $e$ be the identity of the group. For all $g\in G$, we have $$ge=eg=g,$$ so $C(e)=G$ is non-abelian.
A: The centralizer of an element of a group is not abelian in general; $C(a)$ means the largest subgroup of $G$ which its element commutes with a fixed element $a$.
If $a$ is an element of center then $C(a) = G$. So as a counterexample, set $G = S_3$ and $a = 1$ is fine.
