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  "body": "\n## Statement\n\nLet $G$ be a finite [[group|G5S1M9]] and let $p$ be the smallest prime number that divides $|G|$. Let $H$ be a [[subgroup|0X1JXT47GA]] of $G$ of [[index|0DTXBQPA2S]] $p$.\nThen $H$ is [[normal|0CB0DX5BVV]] in $G$.\n\n## Remarks\n\nFor $p = 2$ this is the fact that [[subgroups of index two are normal|0V547YQRMV]].\nThe proof lets $G$ act on the $p$ left cosets of $H$. The kernel $K$ of this action lies in $H$, and $[G : K]$ divides both $|G|$ and $p!$.\nSince $p$ is the smallest prime divisor of $|G|$, this forces $[G : K] = p$, so $H = K$ is normal.\n\n## Explanations\n\n### L1 human\n\nIndex $2$ is not the only small index that forces normality.\nIf the index is the smallest prime that divides the order of the group, the subgroup is also normal.\nFor example, a subgroup of index $3$ in a group of order $75 = 3 \\cdot 5^2$ is normal.\n",
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