§ McKay's proof of Cauchy's theorem for groups [TODO ]

• In a group, if $gh = 1$ then $hg = 1$. Prove this by writing $hg = hg (h h^{-1}) = h(gh)h^{-1} = h \cdot 1 \cdot h^{-1} = 1$.
• We can interpret this as follows: in the multiplication table of a group, firstly, each row contains exactly one $1$.
• Also, when $g \neq h$ (ie, we are off the main diagonal of the multiplication table), each $gh = 1$ has a "cyclic permutation solution" $hg = 1$.
• If the group as even order, then there are even number of $1$s on the main diagonal.
• Thus, the number of solutions to $x^2 = 2$ for $x \in G$ is even, since each solution has another paired with it.
• Let's generalize from pairs to
• Reference