Let G be a finite group
a) Show that the number of elements x of G such that x^3=e is odd
b) Show that the number of elements x of G such that x^2 does not equal e is even
Let $\displaystyle X = \{ x\in G | x^3 = e\}$. Now if $\displaystyle x\not = e$ is in $\displaystyle X$ then $\displaystyle x^3 = e$ so $\displaystyle x$ has order $\displaystyle 3$. Therefore, $\displaystyle x\not = x^2$. But $\displaystyle (x^2)^3 = (x^3)^2 = e$. Thus, $\displaystyle x^2 \in X$. Let $\displaystyle X = \{ e,x_1,x_2,....,x_n,x_{n+1} \}$. Write them in such a way so that $\displaystyle x_1 = x_2^2,....,x_n=x_{n+1}^2$. Therefore, all $\displaystyle x_i$ can be paired with $\displaystyle x_{i+1}$. Thus, there are an even number (zero is allowed) of elements which have order three. But we also include $\displaystyle e$ and this makes the total number of elements odd.
Similar to above.b) Show that the number of elements x of G such that x^2 does not equal e is even