Answer to Question #203298 in Abstract Algebra for Raghav

Question

Let G be a finite group. Show that the number of elements g of G such that g³= e is

odd, where e is the identity of G.

Answer 1

Let $G$ be a finite group. Let us show that the number of elements $g\in G$ such that $g^3= e$ is

odd, where $e$ is the identity of $G$. If $g^3=e$, then $(g^{-1})^3=g^{-3}=(g^3)^{-1}=e^{-1}=e$. Let us find all $g\in G$ such that $g^3=e$ and $g^{-1}=g.$ In this case $g^2=e,$ and hence $g=ge=gg^2=g^3=e.$  Therefore, for all non-identity elements $g$ with $g^3=e$ we have that $(g^{-1})^3=e$ and the set $\{g,g^{-1}\}$ is of cardinality 2. Since $e^3=e$ and $|\{g,g^{-1}\}|=2$ for $g\in G\setminus\{e\}$ and $g^3=e$, we conclude that the number of elements $g\in G$ such that $g^3= e$ is equal to  $|\{e\}\cup(\bigcup\limits_{g\in G\setminus\{e\},\ g^3=e}\{g,g^{-1}\})|=1+2k$ (where $k$ is a number of different pairs $\{g,g^{-1}\}$), and hence is odd.