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

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., 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.

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. 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

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 $P$ 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:

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] Since $P$ is Frattini-in-center, the quotient $P/Z(P)$ has exponent dividing $p$. Thus, every $p^{th}$ power becomes trivial in $P/Z(P)$, hence $g^p \in Z(P)$ for all $g \in P$.
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] By Steps (1) and (2), $\tau_0$ is an endomorphism taking values in $Z(P)$. We can write $\tau$ as $g \mapsto (\tau_0(g))^m$. By Fact (2), and since $Z(P)$ is abelian, the map $x \mapsto x^m$ is an endomorphism of $Z(P)$. The composite map $g \mapsto (\tau_0(g))^m$ therefore gives an endomorphism of $P$ taking values in $Z(P)$.
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] Since $p$ and $mp + 1$ are relatively prime, $mp + 1$ is relatively prime to the order of $P$. Thus, fact (4) applies.
6	The map $\sigma := g \mapsto g^{mp+1}$ is an automorphism of $P$.			Steps (4), (5)	Step-combination-direct.