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For an odd prime $p$ , the $p$ -th power map is a homomorphism on a finite group whose derived subgroup is central of exponent dividing $p$

Statement

Let $p$ be an odd prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and let $G$ be a finite group with $[G, G] \leq Z (G)$ whose derived subgroup has exponent dividing $p$ (The exponent of a finite group, Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ , The center $Z (G)$ of a group). Then

$(x y)^{p} = x^{p} y^{p} for all x, y \in G,$

so $x \mapsto x^{p}$ is a group homomorphism from $G$ to $G$ .

Facts & Assumptions

Given: An odd prime $p$ and a finite group $G$ with $[G, G] \leq Z (G)$ and $\exp ([G, G])$ dividing $p$ ; elements $x, y \in G$ .

[F1]

For a finite group $H$ , $\exp (H) = \min {n \in N : n > 0 and g^{n} = e for every g \in H}$ (The exponent of a finite group).

[F2]

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

[F3]

$2! = 2$ and $n^{\underline{2}} = n (n - 1)$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[L1]

If $[G, G] \leq Z (G)$ then $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$ for every $n \in N$ (In a group with central derived subgroup, $(x y)^{n} = [y, x]^{(\binom{n}{2})} x^{n} y^{n}$ ).

[L2]

For $k \leq n$ one has $(\binom{n}{k}) \cdot k! = n^{\underline{k}}$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

[L3]

For all $m, n \in Z$ one has $(g^{m})^{n} = g^{m n}$ (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 · direct

1.1

Taking $n = p$ in the product formula gives $(x y)^{p} = [y, x]^{(\binom{p}{2})} x^{p} y^{p}$ .

1.2

Since $p$ is odd and $p \geq 3$ , the closed formula at $n = p$ and $k = 2$ gives $(\binom{p}{2}) \cdot 2 = p (p - 1)$ , and writing $p - 1 = 2 m$ with $m \in N$ turns this into $(\binom{p}{2}) \cdot 2 = 2 p m$ , so $(\binom{p}{2}) = p m$ .

F2F3L2algebra

1.3

The element $c = [y, x]$ lies in $[G, G]$ , and $\exp ([G, G])$ divides $p$ , say $p = \exp ([G, G]) \cdot t$ ; hence $c^{p} = (c^{\exp ([G, G])})^{t} = e$ .

F1L3algebra

2.1

Combining, $[y, x]^{(\binom{p}{2})} = (c^{p})^{m} = e$ , so step 1.1 reads $(x y)^{p} = x^{p} y^{p}$ ; as this holds for all $x, y \in G$ , the map $x \mapsto x^{p}$ is a homomorphism.

step 1.1step 1.2step 1.3algebra∎

Remarks

Only the oddness of $p$ is used, in step 1.2; primality enters through the hypothesis on the derived subgroup rather than through the arithmetic. At $p = 2$ the conclusion fails at the first step: $(\binom{2}{2}) = 1$ , so $(x y)^{2} = [y, x] x^{2} y^{2}$ and the commutator factor survives.

Depends on

Used by

Dependency tree · two levels

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Sources

D. A. Craven, The Theory of p-Groups, §3.1, proof of Theorem 3.14 (standard reference, not scraped)
M. van Beek, Topics in Finite p-Groups, §2.3 (standard reference, not scraped)

For an odd prime pp, the pp-th power map is a homomorphism on a finite group whose derived subgroup is central of exponent dividing pp