How can we conclude that $2p\nmid s_{2p}$? 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 $2p\nmid s_{2p}$?
 A: 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 $p).$
