Let $G$ be a finite group. If $x$ is the unique involution of $G$, then $x \in Z(G)$.

Asked Jul 15 '22 at 10:02

Active Jul 15 '22 at 13:28

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My attempt: let $y \in G$ be of even order then $x \in \langle y\rangle$ and $x \in C_G(y)$

If $y$ is of odd order then $|xy|=|x|\times|y|$ and the element $xy$ is of even order. By the previous paragraph we get $x \in C_G(xy)$ i.e $x(xy)=(xy)x$,

yields $xy=yx$ (using cancellation law)

then $x\in C_G(y)$ for every $y\in G$

Finally $x\in Z(G)$

Is my attempt right?

edited Jul 15 '22 at 13:28

Shaun

44,997

asked Jul 15 '22 at 10:02

S.O

I cannot follow anything after "If $y$ is of odd order". – ancient mathematician Jul 15 '22 at 10:07
@LostinSpace yes it answers and it's easier than the way I used but is my attempt right or missing some details? – S.O Jul 15 '22 at 10:12
@ancientmathematician I have added some details I hope it helps – S.O Jul 15 '22 at 10:18
2

Every conjugate of $x$ is also an involution, hence equal to $x$, which means that $x$ is in the center. – Geoffrey Trang Jul 15 '22 at 13:36

Let $G$ be a finite group. If $x$ is the unique involution of $G$, then $x \in Z(G)$.

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