Let $G$ be a finite group, and suppose $p$ is a prime divisor of $|G|$ . Consider the set $X$ of all $p$ -tuples $(x_1, \ldots, x_p)$ for which $x_1\cdots x_p = 1$ . Note that $|X| = |G|^{p-1}$ is a multiple of $p$ . There is a natural group action of the cyclic group $\mathbb{Z}/p\mathbb{Z}$ on $X$ under which $m \in \mathbb{Z}/p\mathbb{Z}$ sends the tuple $(x_1, \ldots, x_p)$ to $(x_{m+1}, \ldots, x_p, x_1, \ldots, x_m)$ . By the Orbit-Stabilizer Theorem, each orbit contains exactly $1$ or $p$ tuples. Since $(1,\ldots, 1)$ has an orbit of cardinality $1$ , and the orbits partition $X$ , the cardinality of which is divisible by $p$ , there must exist at least one other tuple $(x_1,\ldots, x_p)$ which is left fixed by every element of $\mathbb{Z}/p\mathbb{Z}$ . For this tuple we have $x_1 = \ldots = x_p$ , and so $x_1^p=x_1\cdots x_p=1$ , and $x_1$ is therefore an element of order $p$ .

References

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James H. McKay. Another Proof of Cauchy's Group Theorem, American Math. Monthly, 66 (1959), p119.