Let $s_{2p}$ be the number of elements of order $2p$ in finite group $G$ and let $x$ be an element of order $2p$ in $G$. We can write $s_{2p}=\sum_{o(x)=2p}|x^G|$, where these conjugacy classes are distinct. How can we conclude that $2p\nmid s_{2p}$?
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1$\begingroup$ If $p=2$ and $G = C_2 \times C_4$, then $s_4 = 4$. $\endgroup$– S. Carnahan ♦Jan 2, 2015 at 8:02
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$\begingroup$ sorry I forgot to say that $p$ is odd prime. $\endgroup$– ShukranJan 2, 2015 at 8:11
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$\begingroup$ In fact, I want to know how we can conclude $2p\nmid\sum_{o(x)=2p}|x^G|$? $\endgroup$– ShukranJan 2, 2015 at 8:14
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$\begingroup$ You could look at "On a Theorem of Frobenius: solutions of $x^n = 1$ in finite groups" by Isaacs and Robinson. On the way to proving Frobenius Theorem, they show that $\phi(n) \mid s_{n}$ for all $n$, but some of the methods and references might be of use. I assume you want to exclude the case that there are no elements of order $2p$ in $G$, in which case $2p \mid 0$? (E.g. elements of order 6 in $A_{4}$) $\endgroup$– Padraig Ó CatháinJan 2, 2015 at 9:54
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The claim is false in general: the correct statement (when $p$ is an odd prime divisor of $|G|$) is that the number of elements of order $2p$ in $G$ is a multiple of $2p$ if and only if the number of involutions in $C_{G}(P)$ is a multiple of $p,$ where $P \in {\rm Syl}_{p}(G).$ This covers the case that there are no elements of order $2p$.
Suppose that $G$ does contain some element of order $2p$ and that $p$ is odd. Let $P$ be a Sylow $p$-subgroup of $G$, so $|P|$ is divisible by $p.$ Now $P$ permutes the elements of order $2p$ in $G$ by conjugation. Furthermore, whenever $y$ has order $2p,$ then $y$ and $y^{-1}$ are in different $P$-orbits (since no non-identity element of $P$ inverts any element of order greater than $2$), but the orbits of $y$ and $y^{-1}$ have the same length. Hence counting (mod $2p$), we need only concern ourselves with counting $P$ orbits of length $1$. An element $y$ is an orbit of length $1$ if and only if its $2$-part $y_{2}$ is an involution in $C_{G}(P)$ and its $p$-part $y_{p}$ is an element of order $p$ in $Z(P).$
Working (mod $2p$), the number we want is $mn$ where $m$ is the number of involutions in $C_{G}(P)$ and $n$ is the number of elements of order $p$ in $Z(P)$. Now the elements of order $p$ in $Z(P)$, together with the identity, form a subgroup of $p$-power order, so the number of elements of order $p$ in $Z(P)$ is prime to $p$ (and is also even). We are reduced to checking whether or not the number of involutions in $C_{G}(P)$ is prime to $p$. This number can be divisible by $p$: for example, let $G = S \times P$, where $S$ is elementary Abelian of order $8$ and $P$ is cyclic of order $7.$ There are $7$ involutions in $C_{G}(P)$ (and $42$ elements of order $14$ in $G$- $42$ is indeed a multiple of $14$). But we have established the condition stated at the beginning of this answer (in slightly more precise form): for if the number of involutions in $C_{G}(P)$ is a multiple of $p,$ then the number of elements of order $2p$ in $G$ is a multiple of $2p.$ If the number of involutions in $C_{G}(P)$ is not divisible by $p,$ then the number of elements of order $2p$ in $G$ is even, but is not a multiple of $p$.
Late remark: It is true, however, that if $C_{G}(P)$ has a unique conjugacy class of involutions (or, more generally, if $N_{G}(P)$ transitively permutes the involutions of $C_{G}(P)$ by conjugation), then the number of elements of order $2p$ in $G$ is not divisible by $2p$ ( for in that case, the number of involutions in $C_{G}(P)$ is definitely prime to $p).$
answered Jan 2, 2015 at 12:30
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$\begingroup$ 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 $2p$? $\endgroup$– ShukranJan 2, 2015 at 13:35
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$\begingroup$ 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 $2p$. $\endgroup$ Jan 2, 2015 at 13:42
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$\begingroup$ 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. $\endgroup$– ShukranJan 2, 2015 at 13:54
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$\begingroup$ Whether or not $P$ is cyclic makes no difference in the original question. $\endgroup$ Jan 2, 2015 at 14:01
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$\begingroup$ 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). $\endgroup$ Jan 2, 2015 at 14:18