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  "body": "\n## Statement\n\nLet $p$ be a prime and consider operations on $H^*(X; \\F_p)$ for spaces $X$.\nAll [[binomial coefficients|092GEF38YD]] below are read modulo $p$, and $\\binom{m}{k} = 0$ when $k \u003c 0$ or $k \u003e m$.\n\n1. For $p = 2$, the [[Steenrod squares|VR7RHP41M0]] satisfy, for all integers $a, b \u003e 0$ with $a \u003c 2b$,\n$$Sq^a Sq^b = \\sum_{j=0}^{\\floor{a/2}} \\binom{b - 1 - j}{a - 2j} Sq^{a + b - j} Sq^j.$$\n2. For $p$ odd, the [[Steenrod reduced powers|EZYHSE1E8P]] satisfy, for all integers $a, b \u003e 0$ with $a \u003c pb$,\n$$P^a P^b = \\sum_{j=0}^{\\floor{a/p}} (-1)^{a + j} \\binom{(p-1)(b - j) - 1}{a - pj} P^{a + b - j} P^j.$$\n3. For $p$ odd, with $\\beta$ the [[Bockstein homomorphism|H7PCM776GR]] of $0 \\to \\Z/p \\to \\Z/p^2 \\to \\Z/p \\to 0$, for all integers $a, b \u003e 0$ with $a \\le pb$,\n$$P^a \\beta P^b = \\sum_{j=0}^{\\floor{a/p}} (-1)^{a + j} \\binom{(p-1)(b - j)}{a - pj} \\beta P^{a + b - j} P^j + \\sum_{j=0}^{\\floor{(a-1)/p}} (-1)^{a + j + 1} \\binom{(p-1)(b - j) - 1}{a - pj - 1} P^{a + b - j} \\beta P^j.$$\n\n## Remarks\n\nOne proof computes the operations on the cohomology of a product of Eilenberg–MacLane spaces, or of symmetric products, where all compositions are visible.\nAnother proof packages the operations into a generating function and uses the Cartan formula.\n\nExample for $p = 2$: taking $a = b = 1$ gives $Sq^1 Sq^1 = \\binom{0}{1} Sq^2 = 0$.\nTaking $a = 1, b = 2$ gives $Sq^1 Sq^2 = Sq^3$.\nTaking $a = b = 2$ gives $Sq^2 Sq^2 = Sq^3 Sq^1$.\n\nThe condition $a \u003c 2b$ is needed: when $a \\ge 2b$ the composite $Sq^a Sq^b$ is already admissible and is not rewritten.\nThe relations, together with $Sq^0 = 1$ (and $P^0 = 1$, $\\beta^2 = 0$ for odd $p$), give all relations among composites of these operations.\nSo the mod $p$ Steenrod algebra is the free associative algebra on the operations modulo these relations.\n\n## Explanations\n\n### L1 human\n\nSteenrod operations can be composed, and some composites are equal to sums of other composites.\nThe Adem relations list exactly which such equalities hold.\nThey let one rewrite any composite as a sum of composites of a standard form, called admissible.\n\n### L2 human\n\nThe Adem relations make the Steenrod algebra computable: the admissible monomials form a basis, which is the Cartan–Serre basis.\nThey show that $Sq^n$ is decomposable unless $n$ is a power of $2$, so the algebra is generated by the $Sq^{2^k}$.\nThis fact is the starting point of the Hopf invariant one problem and of the Adams spectral sequence.\n",
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