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Having found that a group G has a normal Sylow 2-subgroup P, how do I find $C_P(g_i)$, where $g_i$ is a conjugacy class representative?

I have the character table, and have previously found $|C_G(g_i)|$ and shown that $G/[G,G]$ is cyclic, if those results are relevant.

Thanks in advance!

Edit: $g_i$ is the cyclic group of order 5 $\left\{1, a, a^{2}, a^3, a^4 \right\}$ where $a^5 = 1$.

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Well there is a result in group theory (maybe you did it proving the 1st and/or 2nd Sylow theorem). If $G$ is a group, $S$ a $p$-Sylow of $G$ and $H$ a subgroup of $G$ then there exists $g\in G$ such that $gSg^{-1}\cap H$ is a $p$-Sylow.

Now in your situation $G$ is $G$, $S$ is $P$ and $H$ is $C_G(g_i)$. Then $C_G(g_i)\cap gPg^{-1}$ (for some $g$) is a $2$-Sylow of $C_G(g_i)$. If you do not have more information, I don't think you can conclude. If you mean that the group $G$ has a $\underline{unique}$ Sylow $2$-Subgroup then $gPg^{-1}=P$ and then :

$$C_P(g_i)=C_G(g_i)\cap P=\text{ the unique }2\text{-Sylow of }C_G(g_i)$$

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  • $\begingroup$ Thanks! Yes it is a normal Sylow 2-subgroup, so unique. Sorry I should have put that originally. $\endgroup$
    – Ziggy
    Apr 23, 2015 at 16:32
  • $\begingroup$ Perhaps I've completely misunderstood something then - with $g_i$ as given above, I got that $|C_G(g_i)| = 5$ - meaning there isn't a Sylow 2-subgroup? $\endgroup$
    – Ziggy
    Apr 23, 2015 at 16:35
  • $\begingroup$ Indeed but if the cardinal is $5$ then you can say that $C_P(g_i)$ is trivial using Lagrange's theorem and you don't need what I have done. $\endgroup$
    – Clément Guérin
    Apr 23, 2015 at 16:54
  • $\begingroup$ Thank you for your help! :) $\endgroup$
    – Ziggy
    Apr 24, 2015 at 10:50

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