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?
Best Answer
To see that $n_3$ is even, note that each element $a\in G$ of order $3$ is associated with a subgroup $\{e,a,a^{-1}\}$, and that there are exactly two elements of order $3$ corresponding to each such subgroup.
To see that $o(G)-n_2$ is odd, do something similar.For each $a\in G$ not of order $2$, $a^{-1}$ is not of order $2$ as well, and is thus distinct from $a$ except in the single case $a=e$.