Must the centralizer of an element of a group be abelian?

Solution 1:

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.

Solution 2:

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.