Alternate proof of Cauchy’s Theorem

Let G be a finite group and let p be a prime dividing |G|. Let \mathcal{S} denote the set of all p-tuples of elements of G whose product is 1. That is, \mathcal{S} = \{ (x_1, x_2, \ldots, x_p) \ |\ x_i \in G\ \mathrm{and}\ x_1x_2 \ldots x_p = 1 \}. Now define a relation \sim on \mathcal{S} as follows: \alpha \sim \beta if and only if \beta is a cyclic permutation of \alpha.

  1. Prove that \mathcal{S} has |G|^{p-1} elements, hence has order divisible by p.
  2. Prove that a cyclic permutation of an element of \mathcal{S} is again in \mathcal{S}.
  3. Prove that \sim is an equivalence relation.
  4. Prove that an equivalence class contains a single element if and only if it is of the form (x,x,\ldots,x) for some x with x^p = 1.
  5. Prove that every equivalence class has order 1 or p. Deduce that |G|^{p-1} = k+pd, where k is the number of classes of size 1 and d is the number of classes of size p.
  6. Noting that \{(1,1,\ldots,1)\} is a class of size 1, deduce that there exists a nonidentity element x \in G such that x^p = 1.

  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}.
    1. Every p-tuple is trivially a cyclic permutation of itself, so that \sim is reflexive.
    2. If \sigma is a cyclic permutation of \tau, say by k slots, then \tau may be recovered by cyclically permuting \sigma p-k times.
    3. 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.

  3. (\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.

  4. 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 mod 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.
  5. 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.
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