Let \(G\) be a finite group, let \(n > 0\) be an integer. Let \(A\) be the set of \(n\)-tuples \((x_1, \ldots, x_n)\) of \(G\) such that \(x_1x_2 \ldots x_n = 1\).
(a) Show that \(\mathbb{Z}/n\mathbb{Z}\) acts on \(A\) according to the formula \(\overline{k} \cdot (x_1, \ldots, x_n) = (x_{1+k}, \ldots, x_{n+k})\), where indices are taken mod \(n\).
(b) Suppose \(n = p\) is a prime dividing \(\left|G\right|\). Show that \(p\) divides the number of size 1 orbits, and deduce that the number of elements of order \(p\) in \(G\) is congruent to -1 mod \(p\).
(a) Notice that if \(x_1 \ldots x_n = (x_1 \ldots x_k)(x_{k+1} \ldots x_n) = 1\) then \(x_{1+k} \ldots x_{n+k} = (x_{1+k} \ldots x_n)(x_{n+1} \ldots x_{n+k}) = (x_1 \ldots x_k)(x_{k+1} \ldots x_n) = 1\).
(b) The size of any orbit must divide \(\left|\mathbb{Z}/p\mathbb{Z}\right| = p\), hence the only possibilities are 1 and \(p\). Let the number of size 1 and size \(p\) orbits be \(n_1\) and \(n_p\). The number of possible \(p\)-tuples is easily checked to be \(\left|G\right|^{p-1}\), so we have \(n_1 + pn_p = \left|G\right|^{p-1}\). Since \(p \mid pn_p\) and \(p \mid \left|G\right|^{p-1}\), it follows that \(p \mid n_1\).
If \(\{(x_1, \ldots, x_n)\}\) is a size 1 orbit, then we have \(x_1 = \ldots = x_n\) and \({x_1}^p = 1\). Thus \(x_1\) is either 1 or an order \(p\) element. Eliminating the former possibility yields the congruence relation.
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