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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 ?

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  • $\begingroup$ 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? $\endgroup$ – user87543 Jun 22 '14 at 5:37
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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.
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  • $\begingroup$ Thank you Taladris, your solution is too simple. $\endgroup$ – user117741 Jun 23 '14 at 11:12
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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.

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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.

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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.

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