LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

If $∣ G ∣ = p^{a} m$ with $p ∤ m$ , then $v_{p} (\binom{p^{a} m}{p^{a}}) = 0$

Statement

Let $p$ be prime and let $a, m \in N$ satisfy $p ∤ m$ . Then $v_{p} ⁣ ((\binom{p^{a} m}{p^{a}})) = 0.$ The valuation is applied only to nonzero integers. See The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a prime $p$ and a nonzero integer $x$ , the $p$ -adic valuation is $v_{p} (x) : = \max {k \in N : p^{k} ∣ x};$ the value $v_{p} (0)$ is undefined. (The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ ).

[L2]

For a prime $p$ and nonzero integers $x, y$ , the product $x y$ is nonzero and $v_{p} (x y) = v_{p} (x) + v_{p} (y) .$ ( $v_{p} (a b) = v_{p} (a) + v_{p} (b)$ for nonzero integers $a, b$ , and $v_{p} (a + b) \geq \min {v_{p} (a), v_{p} (b)}$ whenever $a$ , $b$ and $a + b$ are all nonzero).

[L3]

For a finite set $A$ and $k \in N$ put $[A]^{k} : = {S \subseteq A : ∣ S ∣ = k},$ the set of $k$ -element subsets of $A$ . Every $S \subseteq A$ is finite (thm-subset-of-a-finite-set), so the condition $∣ S ∣ = k$ makes sense for every subset. (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L4]

Let $n, k \in N$ with $k \leq n$ . Then $(\binom{n}{k}) k! (n - k)! = n!,$ and consequently $(\binom{n}{k}) k! = n^{\underline{k}}$ . ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

[L5]

The factorial and falling factorial satisfy $0! = 1, (n + 1)! = n! (n + 1), n^{\underline{0}} = 1, n^{\underline{k + 1}} = n^{\underline{k}} (n - k),$ and for $k \leq n$ , $n^{\underline{k}}$ is the product $n (n - 1) \dots (n - k + 1)$ . (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

Proof

technique · direct

1.1L1L2L3L4L5givenalgebra

We apply the valuation to the identity $(\binom{p^{a} m}{p^{a}}) (p^{a})! = (p^{a} m)^{\underline{p^{a}}}$ .

2.1step 1.1givenalgebra

If $0 < j < p^{a}$ , write $j = p^{r} u$ with $p ∤ u$ ; then $r < a$ and $p^{a} m - j = p^{r} (p^{a - r} m - u)$ , whose parenthesized factor is prime to $p$ . Hence $v_{p} (p^{a} m - j) = v_{p} (j)$ . The factor $p^{a} m$ contributes $a + v_{p} (m) = a$ , exactly matching the contribution of $p^{a}$ in $(p^{a})!$ , so all valuations cancel.

3.1step 2.1givenalgebra∎

If $a = 0$ , the binomial coefficient is $(\binom{m}{1}) = m$ , which is nonzero and prime to $p$ . No valuation of zero occurs in either case. This proves the stated claim.

Depends on

Dependency tree · 2 levels

Sources

Keith Conrad, Consequences of the Sylow Theorems, Sections 1-5

If ∣G∣=pam∣G∣=pam with p∤mp∤m, then vp(pampa)=0vp​(papam​)=0