Let $G$ be a finite group and $n_k$ the number of elements of order $k$ in $G$. Show that $n_3$ is even and $o(G) - n_2$ is odd.
By Lagrange's Theorem, if $k$ does not divide $o(G)$, there are no elements of order $k$ in $G$. That implies
$3\!\not|\;o(G)\Longrightarrow n_3=0\Longrightarrow n_3\text{ even}$
and
$2\!\not|\;o(G)\Longrightarrow o(G)\text{ odd}\wedge n_2=0\Longrightarrow o(G)-n_2\text{ odd}\;.$
How must I proceed to calculate $n_3\!\!\mod2$ when $3$ divides the order of $G$ and $o(G)-n_2\equiv n_2$ $\!\!\!\mod2$ when $2$ divides it?