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## Alternate proof of Cauchy’s Theorem

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.9

Solution:

(1) We prove this equality by attempting to choose an arbitrary element of $\mathcal{S}$. Note that the first $p-1$ elements $x_1,x_2, \ldots, x_{p-1}$ of an arbitrary tuple in $\mathcal{S}$ may be chosen arbitrarily – with $|G|$ choices for each; a total of $|G|^{p-1}$ distinct choices. This having been done, the last entry is determined uniquely, namely $(x_1 x_2 \ldots x_{p-1})^{-1}$. This exhausts all possible ways of choosing an element of $\mathcal{S}$, so that $|\mathcal{S}| = |G|^{p-1}$.

(2) Suppose $(x_1,x_2,\ldots,x_p) \in \mathcal{S}$, and let $(x_k,x_{k+1},\ldots,x_p,x_1,\ldots,x_{k-1})$ be a cyclic permutation. Then $$(x_1x_2\ldots x_{k-1})(x_k x_{k+1}\ldots x_p) = 1,$$ so that $$(x_k x_{k+1}\ldots x_p)(x_1x_2\ldots x_{k-1}) = 1.$$ Thus the cyclic permutation is in $\mathcal{S}$.

(3)

(a) Every $p$-tuple is trivially a cyclic permutation of itself, so that $\sim$ is reflexive.
(b) If $\sigma$ is a cyclic permutation of $\tau$, say by $k$ slots, then $\tau$ may be recovered by cyclically permuting $\sigma$ $p-k$ times.
(c) If $\sigma$ is the $k$-th cyclic permutation of $\tau$ and $\theta$ the $\ell-th$ cyclic permutation of $\sigma$, then $\theta$ is the $k+\ell$-th cyclic permutation of $\tau$.

Hence $\sim$ is an equivalence relation.

(4)

($\Rightarrow$) Let $\{ \sigma = (x_1,x_2,\ldots,x_p) \}$ be a $\sim$ equivalence class containing a single element, and let $1 \leq k \leq p$. Now the $k$-th cyclic permutation of $\sigma$ is again $\sigma$, so that $x_k = x_1$ for all such $k$. Thus $\sigma = (x,x,\ldots,x)$.

($\Leftarrow$) Trivial.

(5) Suppose $\sigma \in \mathcal{S}$ has exactly $k$ distinct cyclic permutations; note that $k \leq p$. Now for each $1 \leq i \leq p$, we have $x_i = x_{k+i}$, where indices are taken $\pmod p$. More generally, $x_i = x_{ak+i}$ for each $i$ and $a \in \mathbb{Z}$, with indices taken modulo $p$. Suppose $k$ does not divide $p$; since $p$ is prime, then, $\mathsf{gcd}(k,p) = 1$. We saw previously that $ak+1$ ranges over all residue classes of $\mathbb{Z}/(p)$, so that $$\sigma = (x,x,\ldots,x)$$ for some $x \in G$. Otherwise, $k$ divides $p$, so that $k = 1$ (and $\sigma = (x,x,\ldots,x)$) or $k = p$. Hence every $\sigma \in \mathcal{S}$ is in a $\sim$-equivalence class of size 1 or p. Counting elements of $\mathcal{S}$, then, we have $|G|^{p-1} = k+pd$ where $k$ is the number of size 1 equivalence classes and $d$ the number of size $p$ equivalence classes.

(6) Since $p$ divides $|G|$ and $pd$, we have $p|k$ where $k$ is the number of size 1 equivalence classes. Hence $k > 1$, and there exists at least one size one equivalence class which is not trivial – i.e., has the form $\{ (x,x,\ldots,x) \}$ for some nonidentity $x \in G$. Thus there exists an element $x \in G$ such that $x^p = 1$.