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  "body": "\n## Definition\n\nLet $\\mathbf{A}$ be an [[algebra|QJGPT2W0FT]] with universe $A$.\nLet $\\Delta_A = \\set{(a, a) : a \\in A}$ and $\\nabla_A = A \\times A$.\nA [[congruence|DCSXXT4SER]] $\\theta$ on $\\mathbf{A}$ is a *factor congruence* when there is a congruence $\\theta^*$ on $\\mathbf{A}$ such that:\n\n1. $\\theta \\cap \\theta^* = \\Delta_A$;\n2. $\\theta \\circ \\theta^* = \\nabla_A$, where $(a, c) \\in \\theta \\circ \\theta^*$ means that $(a, b) \\in \\theta$ and $(b, c) \\in \\theta^*$ for some $b \\in A$.\n\nThen $\\theta$ and $\\theta^*$ are called a pair of complementary factor congruences.\n\n## Remarks\n\nFor such a pair, the map $a \\mapsto (a/\\theta, a/\\theta^*)$ is an isomorphism from $\\mathbf{A}$ onto the [[direct product|WJ73XYBNE4]] $\\mathbf{A}/\\theta \\times \\mathbf{A}/\\theta^*$.\nConversely, if $\\mathbf{A} = \\mathbf{B} \\times \\mathbf{C}$, the kernels of the two projections are complementary factor congruences.\nThe conditions say that $\\theta$ and $\\theta^*$ are complements in the [[congruence lattice|REE07CSXHR]] and that they permute.\n\nExample: on the group $\\Z/6\\Z$, take $\\theta$ to be congruence modulo $2$ and $\\theta^*$ congruence modulo $3$.\nAn element congruent to $0$ modulo $2$ and modulo $3$ is $0$, so clause 1 holds.\nFor any $a, c$ there is $b$ with $b \\equiv a \\pmod 2$ and $b \\equiv c \\pmod 3$, by the Chinese remainder theorem, so clause 2 holds.\n\nNon-example: on $\\Z/4\\Z$, congruence modulo $2$ is not a factor congruence.\nThe congruences of $\\Z/4\\Z$ are $\\Delta$, congruence modulo $2$, and $\\nabla$.\nOnly $\\theta^* = \\Delta$ satisfies clause 1, and then $\\theta \\circ \\theta^* = \\theta \\ne \\nabla$.\n\nThe congruences $\\Delta_A$ and $\\nabla_A$ are always factor congruences, complementary to each other.\n",
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    "name": "Factor congruence",
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