Frattini-in-center odd-order p-group implies p-power map is endomorphism

This article states and (possibly) proves a fact that is true for odd-order p-groups: groups of prime power order where the underlying prime is odd. The statement is false, in general, for groups whose order is a power of two.
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Statement

Suppose $p$ is an odd prime, and $P$ is a finite $p$ -group (i.e., a group of prime power order) that is a Frattini-in-center group: the Frattini subgroup of $P$ is contained in its center. Then, the map $x\mapsto x^{p}$ is an endomorphism of $P$ .

Note that this makes it a universal power endomorphism, i.e., an endomorphism described everywhere as raising to a certain power. The endomorphism is nontrivial only if $P$ does not itself have exponent $p$ .

Examples

The smallest non-abelian examples for any odd prime $p$ are the two non-abelian groups of order $p^{3}$ , namely unitriangular matrix group:UT(3,p) (GAP ID $(p^{3},3)$ ) and semidirect product of cyclic group of prime-square order and cyclic group of prime order (GAP ID $(p^{3},4)$ ). Of these two groups, the former has exponent $p$ , so the $p$ -power map is the trivial endomorphism. The latter has exponent $p^{2}$ , so the $p$ -power map is a nontrivial endomorphism.

In the case $p=3$ , these groups are unitriangular matrix group:UT(3,3) and semidirect product of Z9 and Z3 respectively. Both groups have order $3^{3}=27$ .

Related facts

Failure at the prime two

Square map is endomorphism iff abelian, combined with the fact that there exist non-Abelian 2-groups that are Frattini-in-center.

Facts with similar proofs

Omega-1 of odd-order class two p-group has prime exponent

Related facts about power maps

Facts used

Proof

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Given: An odd prime $p$ . A finite $p$ -group $P$ , such that $P/Z(P)$ is elementary Abelian.

To prove: The map $x\mapsto x^{p}$ is an endomorphism of $P$ . Specifically $(xy)^{p}=x^{p}y^{p}$ for any $x,y\in P$ .

Proof:

Step no.	Assertion	Given data used	Facts used	Previous steps used	Explanation
1	The derived subgroup $[P,P]$ is elementary abelian. In particular, $[x,y]^{p}$ is the identity element for any $x,y\in P$ .	$P$ is Frattini-in-center	Fact (1)	--	Follows directly from fact (1).
2	$p$ divides $p(p-1)/2$ .	$p$ is an odd prime.	--	--	Basic properties of divisibility. Note that this breaks down for $p=2$ , because of the $2$ in the denominator.
3	$[x,y]^{p(p-1)/2}$ is the identity element for all $x,y\in P$ .			Steps (1), (2)	By step (1), $[x,y]^{p}$ is the identity element, so the order of $[x,y]$ divides $p$ . Since $p$ divides $p(p-1)/2$ , the order of $[x,y]$ divides $p(p-1)/2$ , so $[x,y]^{p(p-1)/2}$ is the identity element.
4	We have the formula $x^{p}y^{p}=[x,y]^{p(p-1)/2}(xy)^{p}$ for all $x,y\in P$ .	$P$ is Frattini-in-center, and hence class two.	Fact (2)	--	Because $P$ is Frattini-in-center, the quotient by the center is elementary abelian, and hence abelian, so $P$ has class at most two. Thus, we can use fact (2) to get the formula.
5	$x^{p}y^{p}=(xy)^{p}$ for all $x,y\in P$			Steps (3), (4)	This follows directly by plugging in the conclusion of step (3) into step (4).