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  "body": "\n## Definition\n\nLet $G$ be a [[group|G5S1M9]] and let $H$ be a [[subgroup|0X1JXT47GA]] of $G$.\nThe subgroup $H$ is *subnormal* in $G$ when there are subgroups $H_0, H_1, \\dots, H_n$ of $G$ with these properties.\nFirst, $H_0 = H$ and $H_n = G$.\nSecond, $H_{i-1}$ is a [[normal subgroup|0CB0DX5BVV]] of $H_i$ for $1 \\le i \\le n$.\nIn symbols, $H = H_0 \\trianglelefteq H_1 \\trianglelefteq \\cdots \\trianglelefteq H_n = G$.\n\n## Remarks\n\nOne writes $H \\trianglelefteq\\trianglelefteq G$.\nThe least such $n$ is the defect of $H$ in $G$; the subgroups of defect at most $1$ are the normal subgroups.\nExample: in the alternating group $A_4$, let $V = \\{e, (1\\,2)(3\\,4), (1\\,3)(2\\,4), (1\\,4)(2\\,3)\\}$ and $H = \\{e, (1\\,2)(3\\,4)\\}$.\nThen $H \\trianglelefteq V$ because $V$ is abelian, and $V \\trianglelefteq A_4$, so $H$ is subnormal of defect $2$; it is not normal, since conjugation by $(1\\,2\\,3)$ sends $(1\\,2)(3\\,4)$ to $(2\\,3)(1\\,4)$.\nNon-example: in the symmetric group $S_3$, the subgroup $H = \\{e, (1\\,2)\\}$ is not subnormal.\nThe only subgroups of $S_3$ that contain $H$ are $H$ and $S_3$.\nSo a chain from $H$ to $S_3$ must contain a step $H \\trianglelefteq S_3$, but $H$ is not normal in $S_3$.\nIn a [[nilpotent group|0Q7MMQHC1P]] every subgroup is subnormal.\n\n## Explanations\n\n### L1 human\n\nA subgroup is subnormal when you can climb from it to the whole group by a ladder in which each rung is normal in the next one.\n",
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    "name": "Subnormal subgroup",
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