Proof in Group Theory Show that if $G$ is a finite group with identity $e$ and with an even number of elements, then there is an $a \neq e$ in $G$, such that $a \cdot a = e$.
I read the solutions here http://noether.uoregon.edu/~tingey/fall02/444/hw2.pdf
Why do they say $D = \{a, a^\prime\}$? Isn't $D$ not a group? There is no identity and if they include the identity they get 3 elements, which means $|D| = 3 = $ odd.
 A: Consider the relation on $G$ given by $g\equiv h\iff g\in\{\ h\ , h^{-1}\ \}$. It is easy to see that this is symmetric, reflexive, and transitive, and so an equivalence relation with equivalence classes $\{\ h\ ,\ h^{-1}\ \}$. The equivalence class of the identity $e$ of $G$ is $\{\ e\ \}$ containing only one element, and all equivalence classes have at most two elements. Since the order of $G$ is even, at least one equivalence class besides $\{\ e\ \}$ must have only one element, and that element is its own inverse. 
A: You're right that $D$ is not a subgroup of $G$, but they don't claim that it is.
They're also not saying that $D = \{a, a^{-1}\}$, but rather that the elements of $D$ appear in pairs -- if $a \in S$, then $(a^{-1})^2 = a^{-2} = (a^2)^{-1} \neq e$.
Go back through the proof and see if it makes sense now.
A: let $e,a_1,a_2,...,a_n$ be the elements of the group since the number of these elements is even we get the number of these elemnts  $a_1,a_2,...,a_n$ is odd, now start putting every element with its inverse in a set lets say $\{a_i,a_j\}$ since the number is odd you will be left with one element call it $a$ and you have $a.a=e$.
