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  "body": "\n## Proof\n\nWe reduce the class equation modulo $p$. Let $e$ be the identity of $G$.\n\n1. First, $\\abs{G} = p^r$ with $r \\geq 1$. Let $q$ be a prime that divides $\\abs{G}$; one exists since $\\abs{G} \u003e 1$. By [[Cauchy's theorem|0JM7PA288B]], some element of $G$ has order $q$.\n2. Since $G$ is a [[$p$-group|06DAH3VS6M]], that order is a power of $p$. As $q$ is prime, $q = p$.\n3. So $p$ is the only prime factor of $\\abs{G}$, and $\\abs{G} = p^r$ by the [[fundamental theorem of arithmetic|084HH1NDCB]]. Since $\\abs{G} \u003e 1$, $r \\geq 1$.\n4. Each positive divisor $d$ of $p^r$ is a power of $p$. Indeed, a prime factor of $d$ divides $p^r$. So it divides $p$ by [[Euclid's lemma|051VYKA1EX]], and it equals $p$.\n5. By the [[class equation|08BPSC6YTE]], $\\abs{G} = \\abs{Z(G)} + \\sum_{i=1}^{k} [G : C_G(a_i)]$, where $\\operatorname{Cl}(a_1), \\dots, \\operatorname{Cl}(a_k)$ are the conjugacy classes with more than one element.\n6. By the [[class size formula|0VK5V9P8G3]], $[G : C_G(a_i)] = \\abs{\\operatorname{Cl}(a_i)}$, which divides $\\abs{G} = p^r$ and is greater than $1$.\n7. By step 4, each $[G : C_G(a_i)]$ is $p^j$ with $j \\geq 1$, so $p$ divides it. Also $p$ divides $\\abs{G}$, by step 3.\n8. Therefore $p$ divides $\\abs{Z(G)} = \\abs{G} - \\sum_{i} [G : C_G(a_i)]$.\n9. The [[centre|0S7Y9Q990Q]] contains $e$, so $\\abs{Z(G)} \\geq 1$. A positive multiple of $p$ is at least $p \\geq 2$. So $Z(G) \\neq \\set{e}$.\n\n## Idea\n\nIn the class equation every class outside the centre has size a positive power of $p$, and so does the whole group. Modulo $p$ these terms vanish, so the number of central elements is a multiple of $p$. It is not zero, because the identity is central, so it is at least $p$.\n",
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