# How to show that no. of elements $x$ of group $G$ such that $x^3=e$ is odd?

Let G be a finite group G. Then How can I show that no. of elements $x$ of group $G$ such that $x^3=e$ is odd ? I read this question in an Algebra book. Since $e^3=e$, e must be one of those elements. But how to find for non trivial elements ?

• it might be useful if you see that you are being asked to show number of elements of order $3$ are even.... Do you believe that showing this is easy? – user87543 Jun 22 '14 at 5:37

Let $T$ be the set of all elements $x$ in $G$ such that $x^3=e$. Since $e\in T$, we will prove that $S=T\setminus\{e\}$ has an even cardinal.

Details are left to you:

1. If $x$ belongs to $S$, then its inverse $x^{-1}$ is also in $S$.
2. If $x$ is in $S$, then $x\neq x^{-1}$.
3. Every element of $S$ can be paired with another element of $S$, so $S$ has an even number of elements.
• Thank you Taladris, your solution is too simple. – user117741 Jun 23 '14 at 11:12

Every element $x$ such that $x^3 = 1$ is contained in a subgroup $H_x$ of order $3$.

2 different subgroups of order $3$ can intersect only in $1 \in G$, and so the elements such that $x^3$ are $$\lbrace 1 \rbrace \ \ \bigcup \ \ \bigcup_{H_x} \biggl( H_x \setminus \lbrace 1 \rbrace \biggr)$$

In total $2k + 1$ elements.

If $x$ is an element of order 3, then $\langle e, x, x^2 \rangle$ is a cyclic subgroup of order 3 in the group. Notice that $x^2$ also has order 3. For each $x \in G$ of order 3, there is a unique element $x^2=x^{-1}$ which is $\ne x$ and which has order 3. Thus, the number of elements of order 3 in the group is even. Since $e$ also satisfies $x^3=e$, the number of elements $x$ such that $x^3=e$ is odd.

Disclaimer: this is a really worse solution, but I thought it was worth pointing out.

We claim that the number of elements in $$T = \{g\in G: g^3 = e, g\neq e\}$$ is an even number $$N$$. The claim follows from recalling that $$e^3 = e$$, hence the number of elements with trivial cubes is $$N+1$$, an odd number.

In fact, if $$g\in T$$, then $$\{g,g^2,g^4...\}\subset T$$, so, since $$T$$ is finite, there is a minimal number $$N = |T|$$ such that $$g^{2N} \in T$$, and moreover $$g^{2N} = g$$ since $$N$$ is minimal. Suppose by contradiction that $$2^N-1 = 1 \bmod{3} \text{ or } 2^N-1 = 2 \bmod{3}.$$ Then, respectively, $$g^{2N} = g^2$$ and $$g^{2N} = e$$, both which cannot be since $$g \neq g^{-1} = g^2$$ and $$e\notin T$$. Hence $$2^N-1 = 0\bmod{3}.$$

By induction, this is true if $$N$$ is a non-zero even integer, since $$2^2 -1 = 3$$ and $$2^N = 1+3k \implies 2^{N+2} = 1 + 3(k+1).$$ Moreover, it follows that $$2^{2N+1}-1 = 2 \bmod{3}$$, hence $$N$$ is even. This completes the proof.