I have a finite group G, where $Z(G)=1$, and I have an element $g\neq 1$ where $g^{-1} = g$ and $gg=1$. I want to say that $g$ is then only conjugate to itself so that I have a contradiction ($Z(G)$ must then contain at least 2 elements). Can I do this already with the information I have got?
Is it true that if $g^{-1} = g$, then $g$ is only conjugate to itself in a finite group $G$?
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group-theory
finite-groups
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0yup, I just realized :) thanks anyway – 2012-05-05