LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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In a group with central derived subgroup, $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$

Statement

Let $G$ be a group with $[G, G] \leq Z (G)$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ , The center $Z (G)$ of a group) and let $x, y \in G$ . If $[G, G] \leq Z (G)$ then $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$ for every $n \in N$ , where $(\binom{n}{2})$ is the binomial coefficient of The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ and the powers are those of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ .

The commutator on the right is $[y, x]$ , not $[x, y]$ : in the convention $[g, h] = g h g^{- 1} h^{- 1}$ fixed by Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ one has $y x = [y, x] x y$ , and it is that factor which accumulates.

Facts & Assumptions

Given: A group $G$ with $[G, G] \leq Z (G)$ , elements $x, y \in G$ , and $n \in N$ .

[F1]

For $g, h \in G$ the commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F2]

$Z (G) : = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

[F3]

$(\binom{n}{k}) : = ∣ [n]^{k} ∣$ , the number of $k$ -element subsets of $n$ ; in particular $(\binom{n}{1}) = n$ , and $(\binom{n}{k}) = 0$ whenever $k > n$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L1]

If $[G, G] \leq Z (G)$ then $[x y, w] = [x, w] [y, w]$ , $[x, y w] = [x, y] [x, w]$ , and $[x^{n}, y] = [x, y]^{n} = [x, y^{n}]$ for every integer $n$ (Commutator identities in a group whose derived subgroup is central).

[L2]

$(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , with no restriction relating $k$ to $n$ (Pascal's rule $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , and the hockey-stick identity $\sum_{i \leq n} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ ).

[L3]

For all $m, n \in Z$ one has $g^{m + n} = g^{m} g^{n}$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

Proof

technique · induction

1.1

At $n = 0$ the exponent $(\binom{0}{2})$ is zero because $2 > 0$ , and the asserted identity reads $e = e$ ; this is the base case.

F3base

1.2

For every $n \in N$ , Pascal's rule at $k = 1$ gives $(\binom{n + 1}{2}) = (\binom{n}{1}) + (\binom{n}{2}) = n + (\binom{n}{2})$ .

F3L2algebra

1.3

For all $g, h \in G$ one has $[g, h] h g = (g h g^{- 1} h^{- 1}) h g = g h$ , hence $g h = [g, h] h g$ ; applied to the pair $y, x$ this reads $y x = [y, x] x y$ .

F1algebra

1.4

Assume the identity at a given $n \in N$ , that is $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$ .

2.1

Applying step 1.3 to the pair $y^{n}, x$ gives $y^{n} x = [y^{n}, x] x y^{n}$ , and $[y^{n}, x] = [y, x]^{n}$ , so $y^{n} x = [y, x]^{n} x y^{n}$ .

L1step 1.3algebra

2.2

Multiplying the assumption on the right by $x y$ gives $(x y)^{n + 1} = (x y)^{n} (x y) = [y, x]^{(\binom{n}{2})} x^{n} y^{n} x y$ .

step 1.4algebra

3.1

Substituting step 2.1 into step 2.2 and moving the central factor $[y, x]^{n}$ to the front gives $(x y)^{n + 1} = [y, x]^{(\binom{n}{2})} x^{n} [y, x]^{n} x y^{n} y = [y, x]^{(\binom{n}{2}) + n} x^{n + 1} y^{n + 1}$ .

F2L3step 2.1step 2.2algebra

4.1

By step 1.2 the exponent $(\binom{n}{2}) + n$ equals $(\binom{n + 1}{2})$ , so the identity holds at $n + 1$ and therefore at every natural number.

step 1.2step 3.1discharge-induction∎

Remarks

The formula is the reason the $p$ -th power map behaves differently at $p = 2$ : the coefficient $(\binom{2}{2})$ equals $1$ , so the commutator factor survives, whereas for odd $p$ the coefficient $(\binom{p}{2})$ is a multiple of $p$ .

Depends on

Used by

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

D. A. Craven, The Theory of p-Groups, Lemma 2.11 and §3.1 (standard reference, not scraped)
M. van Beek, Topics in Finite p-Groups, §2.3 (standard reference, not scraped)

In a group with central derived subgroup, (xy)n=[y,x](n2)xnyn(xy)n=[y,x](2n​)xnyn