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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.

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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You also need to use the fact that no other element is of order 2 (so no other non-identity element is self-inverse). – M. Vinay Jun 5 '14 at 5:48
up vote 7 down vote accepted

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 $G$.

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Eh apparently MSE is lagging for me, didn't see your answer was posted until after I posted mine. Hope you don't mind if I leave mine up just as a somewhat different perspective. – DanZimm Jun 5 '14 at 6:04
Sure. Feel free. – DeepSea Jun 5 '14 at 6:07

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.

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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$~G$.

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That's very insightful, and related to Eric's answer too. – M. Vinay Jun 5 '14 at 11:57
Isn't that just relations of the form $x^n=e$, I can't think of anything else – Arun Kumar Jun 5 '14 at 13:40
@ArunKumar: No, it can be much more general, since there can be quantified variables that range over the group. The only thing not allowed is constants explicitly designating specific elements (using which one could of course characterise any one element). – Marc van Leeuwen Jun 5 '14 at 14:33

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 $\overset{*}{\implies}$?

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