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

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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.
View other such facts for p-groups|View other such facts for finite groups

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

Facts with similar proofs

Related facts about power maps

Facts used

  1. Frattini-in-center p-group implies derived subgroup is elementary abelian
  2. Formula for powers of product in group of class two

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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^py^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^py^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^py^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).

References

Textbook references