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 $G$?
1
$\begingroup$
group-theory
finite-groups
-
5No. For example $G = S_3$ has three elements squaring to one which are all conjugate to each other and trivial center. – 2012-05-05
-
0Your condition on $g$ is essentially that it has order 2 - so its conjugates will have order 2. Consider elements of order 2 in the symmetric group on three letters (order 6) which has trivial centre. These are all conjugate to each other. – 2012-05-05
-
0I don't know why you think the condition is likely to make $g$ conjugate only to itself... – 2012-05-05
-
0no, I see.. But here is the thing: my group G has order 24, 4 Sylow 3-subgroups and $Z(G)=1$. I consider the action of G on Syl_3(G) by conjugation $\pi: G -> S_{Syl_3(G)}$. Now I want to show that the kernel of $\pi$ cannot consist of two elements (and that it has to be 1 - but this I can conclude if I can just show that the order of the kernel is not 2). I guess I need something to contradict the assumption about $Z(G)=1$. But I am stuck. – 2012-05-05
-
0@Helen: Hint - the kernel of $\pi$ is a normal subgroup of $G$. – 2012-05-05
-
0yup, I just realized :) thanks anyway – 2012-05-05