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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?

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    yup, I just realized :) thanks anyway2012-05-05

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