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I would be grateful if someone could comment on my solution to the following question.

Let $G$ be a finite group. If $g \in G$ and $g \neq e_{G}$, then prove that $|C_{G}(g)| > 1$. [ $C_{g}$ denotes the centraliser of $g$ in $G$. ]

My attempt is that as $C_{G} \leq G$, then $e_{G}$ is in both G and $C_{G}$. So if $g \neq e_{G}$, then $g$ must be a second element in both groups, and so the order of $C_{G}$ must be at least 2. I'm told that there is another (easier) method for reasoning the above???

Mårten W
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Gismho
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  • I don't understand your comment regarding $C_G≤G$. All you need to do is to remark that $e_G,g\in C_G(g)$. – lulu Sep 02 '17 at 13:03
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    Your argument is correct and the easiest possible one: both $e$ and $g$ are guaranteed to commute with $g$. If someone tells you there is an easier way, ignore them. – Randall Sep 02 '17 at 13:09
  • Thank you for prompt responses. The statement $C_{G}\leq G$ was meant to indicate that $C_{G}$ was a subgroup of G. Perhaps I've used the wrong terminology! – Gismho Sep 02 '17 at 15:26

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$\{g^n:n\in \mathbb Z\}\le C_G(g).$ Hence $\mid C_G(g)\mid=1$ implies $\mid \{g^n:n\in \mathbb Z\}\mid =\{e_G\}\Rightarrow g=e_G$ which is not the case.

Supriyo Halder
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