# Powering-injectivity is central extension-closed

## Statement

Suppose $G$ is a group and $H$ is a central subgroup of $G$. Suppose $p$ is a prime number such that:

• $H$ is $p$-powering-injective, i.e., $x \mapsto x^p$ is injective from $H$ to itself.
• The quotient group $G/H$ is $p$-powering-injective, i.e., $x \mapsto x^p$ is injective from $G/H$ to itself.

Then, the whole group $G$ is $p$-powering-injective.

## Proof

Given: A group $G$, a prime number $p$. A central subgroup $H$ of $G$ such that in both $H$ and $G/H$ viewed separately, the map $x \mapsto x^p$ is injective. Two elements $a,b \in G$ with $a^p = b^p$.

To prove: $a = b$.

Proof: Let $\pi:G \to G/H$ be the quotient map.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $\pi(a) = \pi(b)$. $a^p = b^p$, and the map $x \mapsto x^p$ is injective in $G/H$. We have $a^p = b^p$, hence $\pi(a^p) = \pi(b^p)$. Thus, $\pi(a)^p = \pi(b)^p$, so $\pi(a),\pi(b)$ are elements of $G/H$ with the same $p^{th}$ power. Thus, they must be equal.
2 The element $u = ab^{-1}$ is an element of $H$. Step (1) $\pi(u) = \pi(b^{-1}) = \pi(a)\pi(b)^{-1}$ is the identity element of $G/H$, so $u \in H$.
3 $u^p$ is the identity element of $H$. $a^p = b^p$, $H$ is central in $G$. Step (2) We have $a = ub$. Thus, $a^p = (ub)^p$. By the centrality of $H$, we get math>a^p = u^pb^p[/itex]. We also have $a^p = b^p$. This gives $b^p = u^pb^p$, so $u^p$ is the identity element of $G$, and hence also of the subgroup $H$.
4 $u$ is the identity element of $H$. $x \mapsto x^p$ is injective in $H$. Step (3) Step-given direct.
5 $a = b$. Steps (2), (4) Step-combination direct.