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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.

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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.

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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 $G$.

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$~G$.

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.

**Hint:**

$$

y y^{-1} = e \implies yxxy^{-1} = e \implies yxy^{-1} yxy^{-1} = e \implies \left( yxy^{-1} \right)^2 = e \overset{*}{\implies} yxy^{-1} = x

$$

Now the real question is why do we have the implication denoted by $\overset{*}{\implies}$?

If a group contain exacly one element of order 2.then it has exactly one sub group order 2.so it must be normal .so for any Y belongs to G yxy^-1 should be in x .and it shoud be x so from that we say that X must be in centre of G

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