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

Statement
Suppose $$p$$ is an odd prime, and $$P$$ is a finite $$p$$-group (i.e., an fact about::odd-order p-group: a fact about::group of prime power order with an odd prime). Then, if $$P$$ is a fact about::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.

Examples
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

 * Frattini-in-center odd-order p-group implies p-power map is endomorphism
 * Square map is endomorphism iff abelian
 * Inverse map is automorphism iff abelian
 * Cube map is endomorphism iff abelian (if order is not a multiple of 3)
 * nth power map is endomorphism iff abelian (if order is relatively prime to n(n-1))

Facts used

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

Proof
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 $$P$$.

Proof: