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# proof of Cauchy’s theorem

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

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

## Mathematics Subject Classification

20E07*no label found*20D99

*no label found*

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