# 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 $3^3 = 27$.

## 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 $(xy)^p = x^py^p$ 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).