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  "body": "\n## Proof\n\nWe write each element of $G$ as a power of one element times a central element.\n\n1. Let $G$ be a group with identity $e$ and [[centre|0S7Y9Q990Q]] $Z = Z(G)$. Assume that the [[quotient group|0YYWCNTPWR]] $G/Z$ is [[cyclic|0B85Q41RP6]], with generator $tZ$ for some $t \\in G$.\n2. The projection $\\pi \\colon G \\to G/Z$, $\\pi(x) = xZ$, satisfies $\\pi(xy) = (xy)Z = (xZ)(yZ) = \\pi(x) \\pi(y)$. So $\\pi$ is a [[group homomorphism|0HV9Y7WG5R]].\n3. For $k \\ge 0$, $\\pi(t^k) = \\pi(t)^k$ by induction on $k$. The case $k = 0$ holds because [[homomorphisms preserve the identity|08FKQD7NJ3]], and $\\pi(t^{k+1}) = \\pi(t^k) \\pi(t)$.\n4. For $k \u003e 0$, $\\pi(t^{-k}) = \\pi((t^{-1})^k) = \\pi(t^{-1})^k = (\\pi(t)^{-1})^k = \\pi(t)^{-k}$, since homomorphisms preserve inverses.\n5. By steps 3 and 4, $(tZ)^k = t^k Z$ for each integer $k$.\n6. Let $x \\in G$. By the [[definition of $\\langle tZ \\rangle$|0YT586TKXF]] and step 5, $xZ = t^m Z$ for some integer $m$.\n7. Since $e \\in Z$, $x = xe$ lies in the [[left coset|05FD317V2W]] $xZ = t^m Z$. So $x = t^m z_1$ for some $z_1 \\in Z$.\n8. Let $x, y \\in G$. By step 7, $x = t^m z_1$ and $y = t^n z_2$ with integers $m, n$ and $z_1, z_2 \\in Z$.\n9. Since $z_1$ commutes with every element, $xy = t^m z_1 t^n z_2 = t^m t^n z_1 z_2$. By the [[laws of exponents|00DRK90Z6K]], $xy = t^{m+n} z_1 z_2$.\n10. In the same way, $yx = t^n z_2 t^m z_1 = t^n t^m z_2 z_1 = t^{m+n} z_2 z_1$.\n11. Since $z_1 \\in Z$, $z_1 z_2 = z_2 z_1$. By steps 9 and 10, $xy = yx$.\n12. Step 11 holds for all $x, y \\in G$. So $G$ is [[abelian|071DDFVG4J]].\n\n## Idea\n\nModulo the centre, every element is a power of $t$, so each element is $t^m$ times a central element. Powers of the same $t$ commute with each other, and central elements commute with everything. So any two elements commute.\n",
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