The strategy is to let $C_p$ act on a carefully chosen set $X$ of $p$-tuples, show $|X|$ is divisible by $p$, then use the Orbit–Stabiliser theorem to conclude that orbits of size $1$ (other than the trivial tuple) must exist. These correspond to elements of order $p$. **Step 1: Define the [set](/page/Set) and the action.** Let: \begin{align*} X = \{(x_1, x_2, \ldots, x_p) : x_i \in G, \; x_1 x_2 \cdots x_p = e\}. \end{align*} The first $p - 1$ entries may be chosen freely, and $x_p = (x_1 \cdots x_{p-1})^{-1}$ is then determined. So $|X| = |G|^{p-1}$. Since $p \mid |G|$, we have $p \mid |X|$. Let $C_p = \langle h \rangle$ act on $X$ by cyclic permutation: \begin{align*} h^i \cdot (x_1, \ldots, x_p) = (x_{1+i}, x_{2+i}, \ldots, x_{p+i}), \end{align*} where indices are taken modulo $p$. [claim:Cyclic Shift Preserves X] The action maps $X$ to $X$, i.e., if $(x_1, \ldots, x_p) \in X$, then $(x_{1+i}, \ldots, x_{p+i}) \in X$. [/claim] [proof] We need $x_{1+i} x_{2+i} \cdots x_{p+i} = e$ (indices mod $p$). Let $s = x_1 \cdots x_i$. Then: \begin{align*} x_{i+1} \cdots x_p \cdot x_1 \cdots x_i = s^{-1}(x_1 \cdots x_p)s = s^{-1} e \, s = e. \end{align*} [/proof] **Step 2: Analyse orbit sizes.** By the [Orbit-Stabiliser Theorem](/theorems/796), every orbit of $C_p$ acting on $X$ has size dividing $|C_p| = p$. Since $p$ is prime, each orbit has size $1$ or $p$. **Step 3: Count orbits of size $1$.** The orbits partition $X$, so: \begin{align*} |X| = (\text{number of size-1 orbits}) + p \cdot (\text{number of size-}p\text{ orbits}). \end{align*} Since $p \mid |X|$ and $p$ divides the second term, $p$ divides the number of size-$1$ orbits. The tuple $(e, e, \ldots, e)$ is a fixed point (orbit of size $1$), so there is at least one. Since the count of such orbits is divisible by $p \geq 2$, there must be at least $p$ of them, hence at least one non-trivial fixed point. **Step 4: Extract the element of order $p$.** A fixed point $(g, g, \ldots, g) \in X$ satisfies $g^p = e$. If $g \neq e$, then $o(g) \mid p$ by the [Order Division Lemma](/theorems/771), and $o(g) \neq 1$, so $o(g) = p$.