Please Explain the Importance of Equivalence Classes in Classifying Groups of Order $2p$ I'm trying to go through the solutions at this link and I'm not exactly understanding the solution for part (d). The question is:
Suppose that $G$ does not contain an element of order $2p$.  Show that $G$ must contain an element of order 2.
The solution is given as:
Now suppose that every element has order 1 or $p$.  We first show that the following relation is an equivalence relation on the set of non-identity elements of $G:a∼b$ if there is an $n$ so that $a=bn$.  It is reflexive (taking $n= 1$) and transitive (if $b=c^m$ then $a=c^{mn}$).  In the equation $a=b^n$ we must have $n$ relatively prime to $p$ since $a$ is not the identity.  Taking $m$ to be the inverse of $n \mod p$ and raising both sides of $a=b^n$ to the $m$th yields $b=a^m$, so it is symmetric. Each equivalence class under this relation has size $p−1$.  But there are $2p−1$ elements of $G$,which is not divisible by $p−1$.  This provides the desired contradiction.
I really don't understand this solution - how do equivalence classes come into the contradiction?
 A: The big picture idea is to obtain some control over the possible orders in your group $G$ given that $G$ is a group of order $2p$ that does not have any elements of order $2p$. We want to guarantee the existence of an element of order $2$, but to do this we will need to give ourselves something more to work with: a priori certainly $2$ is possible, since it divides the order of the group, but we don't have any theorems guaranteeing the existence of such an element.
This is where the equivalence relation comes in: we assume that there is no element of order $2$, so that every element of the group has order $p$ or is the identity. To derive a contradiction, we should look for some divisibility criterion we can violate. The equivalence relation cooks this up for us.
The idea with the equivalence relation is to collect all the nonidentity group elements in each cyclic subgroup of $G$. Each cyclic subgroup will have $p-1$ nonidentity elements, and by assumption, these sets of size $p-1$ partition the $2p-1$ nonidentity elements of $G$. Now we can observe that we've found our contradiction.
