# Finite implies powering-invariant

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., finite subgroup) must also satisfy the second subgroup property (i.e., powering-invariant subgroup)
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## Statement

Suppose $G$ is a group and $H$ is a finite subgroup of $G$. Then, $H$ is a powering-invariant subgroup of $G$. Explicitly, if $p$ is a prime number such that $G$ is powered over $p$, then $H$ is also powered over $p$.

## Proof

Given: A group $G$, a finite subgroup $H$, a prime number $p$ such that the map $x \mapsto x^p$ is bijective from $G$ to itself.

To prove: The map $x \mapsto x^p$, restricted to $H$, is bijective from $H$ to itself.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The map $x \mapsto x^p$ sends $H$ to within itself. $H$ is a subgroup.
2 The map $x \mapsto x^p$ (restricted to $H$) is injective from $H$ to itself. The map $x \mapsto x^p$ is bijective from $G$ to itself. Step (1) The map is bijective on $G$, hence injective on $G$, hence its restriction to $H$ must also be injective.
3 the map $x \mapsto x^p$ (restricted to $H$) is injective from $H$ to itself. $H$ is finite. Step (2) Since $H$ is finite, any injective map from $H$ to itself must be bijective.