Prove that if a group contains exactly one element of order 2, then that element is in the center of the group.

I'm stuck at this question. Can someone please help me?

Prove that if a group contains exactly one element of order 2, then that element is in the center of the group.

Let $x$ be the element of $G$ which has order 2. Let $y$ be an arbitrary element of $G$. We have to prove that $x \cdot y = y \cdot x$.

Since $x$ has order $2$, \begin{equation} x^2 = e \end{equation} That is, \begin{equation} x^{-1}=x \end{equation}

I don't really know how to proceed. I've tried a number of things, but none of them seem to work.

Solution 1:

Consider the element $z =y^{-1}xy$, we have: $z^2 = (y^{-1}xy)^2 = (y^{-1}xy)(y^{-1}xy) = e$. So: $z = x$, and $y^{-1}xy = x$. So: $xy = yx$. So: $x$ is in the center of $G$.

Solution 2:

More generally, if a group$~G$ contains exactly one element$~x$ having any given property that can be expressed in the language of group theory (in particular without mentioning any specific element of$~G$, other than the identity element$~e$), then $x$ is in the center of$~G$. Namely, any automorphism of$~G$ must send $x$ to an element with the same property, which means it has to fix$~x$. In particular this is the case for inner automorphisms (conjugation by some element of$~G$), and this implies that $x$ is in the centre of$~G$.

Solution 3:

Every element of a conjugacy class has the same order. Since there is only one element of order 2 that element forms a singleton conjugacy class. An element has a singleton conjugacy class iff it is in the center.

These are basic observations once you get to the class equation.