Frattini-in-center odd-order p-group implies (mp plus 1)-power map is automorphism

From Groupprops
Jump to: navigation, search
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


Suppose p is an odd prime, and P is a finite p-group (i.e., an Odd-order p-group (?): a Group of prime power order (?) with an odd prime). Then, if P is a Frattini-in-center group (?), i.e., if \Phi(P) \le Z(P), and m is any integer, then the map g \mapsto g^{mp+1} is an automorphism of P.

Thus, it is a universal power automorphism. In the particular case that P does not have exponent p, this gives a non-identity universal power automorphism.


For any odd prime p, the smallest non-abelian examples are the groups of order p^3. There are two such examples: prime-cube order group:U(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)). The former has exponent p, so the (p+1)- power map is the identity automorphism. The latter has exponent p^2, so that (p+1)-power map is a non-identity universal power automorphism. In fact, this automorphism itself has order p.

For the case p = 3, these groups become prime-cube order group:U(3,3) and semidirect product of Z9 and Z3 respectively, both of order 27.

Related facts

Facts used

  1. Frattini-in-center odd-order p-group implies p-power map is endomorphism
  2. Abelian implies universal power map is endomorphism
  3. (n-1)th power map is endomorphism taking values in the center implies nth power map is endomorphism
  4. kth power map is bijective iff k is relatively prime to the order


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 that is a Frattini-in-center group, i.e., P/Z(P) is elementary abelian, or equivalently, \Phi(P) \le Z(P). m is an integer.

To prove: The map g \mapsto g^{mp + 1} is an automorphism of P.


Step no. Assertion Facts used Given data used Previous steps used Explanation
1 Define \tau as the map g \mapsto g^p. Then \tau is an endomorphism of P. Fact (1) P is Frattini-in-center, and p is odd. -- Fact+Given direct.
2 The image of \tau_0:= g \mapsto g^p is in Z(P). P is Frattini-in-center. [SHOW MORE]
3 The map \tau:g \mapsto g^{mp} is an endomorphism of P taking values in Z(P). Fact (2) Steps (1), (2) [SHOW MORE]
4 The map \sigma:= g \mapsto g^{mp + 1} is an endomorphism of P. Fact (3) Step (3) Fact+Step direct, setting n = mp + 1.
5 The map \sigma := g \mapsto g^{mp+1} is bijective from P to P. Fact (4) [SHOW MORE]
6 The map \sigma := g \mapsto g^{mp+1} is an automorphism of P. Steps (4), (5) Step-combination-direct.