# Centraliser in a subgroup of an element not in that subgroup

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 $h\not \in P$?

• 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. – Derek Holt May 11 '17 at 20:32
• 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 $9$. – 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.
• 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 \}\$. – Riju May 16 '17 at 5:29