Suppose $G, \ast$ is a finite group or order $2p$, where $p$ is an odd prime. I was able to prove that there must be a subgroup of order $p$ denoted by $H$ and that for each element $g \in G$ we have that $g^2 \in H$. The third question I needed to solve regarding this exercise is the number of elements of order $2$ in the group $G$. I think I should be able to use what I found before.
I also know (from a previous exercise) that each abelian group having at least two elements of order 2 has a subgroup of order 4, so for abelian groups, I have at most 1 element of order 2. I also know that the dihedral group of order $2p$ has $p$ elements of order 2. However, I have no idea on how to prove the number of elements more generaly... Any hints? (I only saw the definition of groups, subgroups, cosets and order of groups/subgroups and elements). Thank you in advance.