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Jan 2, 2015 at 16:40 history edited Geoff Robinson CC BY-SA 3.0
very minor typo
Jan 2, 2015 at 14:26 history edited Geoff Robinson CC BY-SA 3.0
elaborated on special case
Jan 2, 2015 at 14:18 comment added Geoff Robinson For example, if $G = E \times F$ where $E$ is elementary Abelian of order $8$ and $F$ is elementary Abelian of order $49,$ the number of elements of order $14$ in $G$ is $7 \times 48$, as my proof in the answer indicates it should be (remember also that the calculation in the answer is only true in general ( mod 2$p$) , as indicated).
Jan 2, 2015 at 14:01 comment added Geoff Robinson Whether or not $P$ is cyclic makes no difference in the original question.
Jan 2, 2015 at 13:54 comment added Shukran We have this Theorem: Let $G$ be a group and $P$ be a cyclic Sylow $p$-subgroup of $G$ of order $p^a$. If there is a prime $r$ such that $p^ar\in \omega(G)$, then $s_{p^ar}=s_r(C_G(P))s_{p^a}$. In particular, $\phi(r)s_{p^a}\mid s_{p^ar}$. So I think that $P$ should be cyclic.
Jan 2, 2015 at 13:42 comment added Geoff Robinson I do not understand what you mean in the question about "if $P$ is cyclic". As for $P$ permuting the elements of order $2p$ by conjugation, I mean that $y \to x^{-1}yx$ for $y$ of order $2p$ and $ x \in P$ gives a permutation action of $P$ on elements of order $2p$.
Jan 2, 2015 at 13:35 comment added Shukran I appreciate you for your answer. I have a question: in statement that you write if $P$ is cyclic? and why do you say that $P$ permutes the elements of order $2p$?
Jan 2, 2015 at 12:59 history edited Geoff Robinson CC BY-SA 3.0
added extra remark
Jan 2, 2015 at 12:51 history edited Geoff Robinson CC BY-SA 3.0
typo
Jan 2, 2015 at 12:44 history edited Geoff Robinson CC BY-SA 3.0
minor adjustment of text
Jan 2, 2015 at 12:36 history edited Geoff Robinson CC BY-SA 3.0
minor rearrangement of text
Jan 2, 2015 at 12:30 history answered Geoff Robinson CC BY-SA 3.0