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  "body": "\n## Proof\n\nWe split $G$ into conjugacy classes and sort the classes by size.\n\n1. The [[conjugacy classes|0Q88KQD17A]] are the orbits of the [[action by conjugation|0P8CY9V437]]. Since [[orbits partition|0ZC7FYE0NZ]] the set, the classes are pairwise disjoint and cover $G$.\n2. Let $a \\in G$. Then $a = e a e^{-1} \\in \\operatorname{Cl}(a)$. So $\\operatorname{Cl}(a) = \\set{a}$ if and only if $g a g^{-1} = a$ for all $g \\in G$.\n3. As $g a g^{-1} = a$ is equivalent to $g a = a g$, step 2 says $\\operatorname{Cl}(a) = \\set{a}$ if and only if $a$ lies in the [[centre|0S7Y9Q990Q]] $Z(G)$.\n4. By step 3, the classes with one element are the sets $\\set{z}$ with $z \\in Z(G)$. Together they contain exactly $\\abs{Z(G)}$ elements.\n5. The classes with more than one element are $\\operatorname{Cl}(a_1), \\dots, \\operatorname{Cl}(a_k)$, by the choice of $a_1, \\dots, a_k$. Since $G$ is finite, step 1 gives $\\abs{G} = \\abs{Z(G)} + \\sum_{i=1}^{k} \\abs{\\operatorname{Cl}(a_i)}$.\n6. By the [[class size formula|0VK5V9P8G3]], $\\abs{\\operatorname{Cl}(a_i)} = [G : C_G(a_i)]$. So $\\abs{G} = \\abs{Z(G)} + \\sum_{i=1}^{k} [G : C_G(a_i)]$.\n\n## Idea\n\nThe conjugacy classes cut the group into disjoint pieces. The pieces with one element are exactly the central elements, and each other piece has the size of the index of a centraliser. Adding up the sizes gives the equation.\n",
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