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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

$$3\!\not|\;o(G)\Longrightarrow n_3=0\Longrightarrow n_3\text{ even}$$


$$2\!\not|\;o(G)\Longrightarrow o(G)\text{ odd}\wedge n_2=0\Longrightarrow o(G)-n_2\text{ odd}\;.$$

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?

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up vote 4 down vote accepted

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 $a=e$.

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HINT for $n_3$: If $g$ is an element of order $3$, so is $g^2\ne g$. There’s no need to look at cases.

HINT for $n_2$: Similarly, if $g$ is not of order $2$ and not the identity, then $g^{-1}\ne g$ is also not of order $2$ and not the identity.

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It known $n_{k}$= $m\phi (k)$, where $m$ is the number of cyclic subgroups of order $k$ in the group $G$ and $\phi $ is the Euler function. If $k>2 $, then $\phi (k)$ is even. Thus always $n_{3}$ is even and $n_{2}$ is odd. Since order of $G$ is even, then $O(G)-n_{2}$ is odd.

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Why is order of $G$ even? o.k. if it has no element of order 2 then order of $G$ is odd. – PAD Jul 15 '12 at 9:24

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