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.
Best Answer
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 $G$.