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  "body": "\n## Statement\n\nLet $G$ be a [[group|G5S1M9]].\nA refinement of a [[subnormal series|0ZZR8D7Y1Z]] of $G$ is a subnormal series of $G$ that contains every term of the given series.\nTwo subnormal series of $G$ are equivalent when they have the same length and their factors agree up to a permutation and isomorphism.\nThen any two subnormal series of $G$ have refinements that are equivalent.\n\n## Remarks\n\nLet the two series be $(G_i)_{0 \\le i \\le n}$ and $(H_j)_{0 \\le j \\le m}$.\nThe proof inserts the groups $G_{i,j} = G_{i-1}(G_i \\cap H_j)$ between $G_{i-1}$ and $G_i$, and the groups $H_{j,i} = H_{j-1}(H_j \\cap G_i)$ between $H_{j-1}$ and $H_j$.\nBoth refined series have length $nm$, and the [[Zassenhaus lemma|21JJ523V95]] gives $G_{i,j}/G_{i,j-1} \\cong H_{j,i}/H_{j,i-1}$ for each pair $(i, j)$.\nFor example, in $\\Z/12\\Z$ the series $0 \\le 6\\Z/12\\Z \\le \\Z/12\\Z$ and $0 \\le 4\\Z/12\\Z \\le \\Z/12\\Z$ refine to $0 \\le 6\\Z/12\\Z \\le 2\\Z/12\\Z \\le \\Z/12\\Z$ and $0 \\le 4\\Z/12\\Z \\le 2\\Z/12\\Z \\le \\Z/12\\Z$.\nBoth refinements have factors of orders $2, 3, 2$ in some order, namely $\\Z/2\\Z$, $\\Z/3\\Z$, $\\Z/2\\Z$ and $\\Z/3\\Z$, $\\Z/2\\Z$, $\\Z/2\\Z$.\n\n## Explanations\n\n### L1 human\n\nA subnormal series cuts a group into layers.\nTwo different ways of cutting may look unrelated.\nThe theorem says that you can always cut each of them further so that both give the same list of layers, up to order.\n\n### L2 human\n\nThe theorem is the main step in the proof of the [[Jordan–Hölder theorem|0E088ZP533]].\nA [[composition series|0XV2A49X6Q]] has no proper refinement other than itself with repeated terms, so two composition series are already equivalent once the repeated terms are removed.\nThe same argument works for groups with operators and for modules, where it gives the uniqueness of composition factors of modules of finite length.\n",
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