Algebra (group theory) Prove, without using Cauchy’s Theorem, that any finite group $G$ of even order contains an element of order two. [Hint: Let $S = \{\,g ∈ G : g \ne g^{−1}\,\}$. Show that $S$ has even number of elements. Argue that every element not in $S$ and not equal to the identity element has order two]. 
My Solution:
Let $S = \{\,g ∈ G : g \ne g^{−1}\,\}$. If $g ∈ S$ then $g^{-1} ∈ S$. This implies that the number of elements in $S$ must be even. Since $g\ne g^{−1}$, we can eliminate all non-identity elements. Then the number of elements of the group is $2n+1$, where $n$ is the number of pairings and $1$ is the identity element $e$. But we stated that $S$ is even. Therefore, this is a contradiction and $G$ has an element of order $2$.

In principle, this is fine - almost. However, you end with "Therefore, this is a contradiction" - but you didn't even start by making an assumption that can be contradicted! Presumably (but you should have explicitly said so!) you wanted to start from a finite group $G$ of even order that does not have an element of order $2$. Also, a bit more carefulness at this step would be in orde: "If $g\in S$ then $g^{-1}\in S$. This implies that the number of elements in $S$m must be even." Are you aware that there exists a subset $S'$ of $G$ that also has the property "If $g\in S'$ then $g^{-1}\in S'$", but has en odd number of elements? 
By the way, you could as well keep this up as a direct proof: Once you show that $|S|$ is even, conclude that $|G\setminus S|$ is also even, contains $e$ and hence at least one second element, $a$ say, and then argue that the order of $a$ is $2$ as was to be shown.
