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  "body": "\n## Proof\n\n1. Let $G$ be an [[abelian group|071DDFVG4J]], and let $G_0 = \\{e\\}$ and $G_1 = G$.\n2. The subgroup $\\{e\\}$ is [[normal|0W77CVBCHQ]] in $G$, because $geg^{-1} = e$ for each $g \\in G$.\n3. So $\\{e\\} = G_0 \\trianglelefteq G_1 = G$ is a [[subnormal series|096HN851YZ]] of $G$ of length $1$.\n4. Its only factor is the [[quotient group|0YYWCNTPWR]] $G/\\{e\\}$. For $a, b \\in G$ we have $(a\\{e\\})(b\\{e\\}) = (ab)\\{e\\} = (ba)\\{e\\} = (b\\{e\\})(a\\{e\\})$, because $G$ is abelian.\n5. So the factor $G/\\{e\\}$ is abelian, and $G$ is [[solvable|0P3KZK0FTV]] by definition.\n\n## Idea\n\nThe shortest possible series, from the trivial group straight to $G$, already works. Its single factor is a copy of $G$, which is abelian.\n",
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