The following is a table of conjugacy class representatives for some group G:

$$ \begin{array}{c|cccccc} &g_1&g_2&g_3&g_4&g_5&g_6\\ \hline |C_G(g_i)|&36&4&9&9&4&4\\ |g_i^G|&1&9&4&4&9&9 \end{array} $$

I'm given that $P$ is a Sylow 3-subgroup, so it must be the union $$P=g_1 \cup g_3^G \cup g_4^G.$$

I'm trying to show that $C_P(g_5)=\{1\}$. Is the following proof correct?

We have $C_P(g_5) \subseteq C_G(g_5)$, so $|C_P(g_5)| \leq 4$. But a centraliser is a normal subgroup, so is a union of conjugacy classes, and must include the indentity. So $C_P(g_5)=\{1\}$.

I'm not sure about the following: I know that $C_G(h)$ is normal in $G$ for all $h$, but what about the $C_P(h)$? In particular, what if $h\not \in P$?

  • $\begingroup$ It is not true in general that centralizers are normal subgroups. Your statement that $C_G(h)$ is normal in $G$ for all $h$ is wrong. $\endgroup$ – Derek Holt May 11 '17 at 20:32
  • $\begingroup$ Note that $C_P(g_5)$ is a subgroup both of $C_G(g_5)$, which has order $4$, and of $P$, which has order $9$. $\endgroup$ – Derek Holt May 11 '17 at 20:33

$C_{P}(g_{5})=C_{G}(g_{5}) \cap P$. P is 3-Sylow so P has order 9. $|C_{G}(g_{5})|=4$ as given. So you cannot have nontrivial intersection.

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  • $\begingroup$ Thanks. Does this use the fact that Sylow p-subgroups for different primes have a trivial intersection... or is there a more general fact? $\endgroup$ – Mark Butler May 15 '17 at 19:48
  • $\begingroup$ It's not about Sylow subgroup. It's just that if $|H|$ and $|K|$ are relatively prime then you cannot have a nontrivial intersection,i.e, $H\cap K=\{\ e \}\ $. $\endgroup$ – Riju May 16 '17 at 5:29

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