Normal of finite index implies quotient-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., normal subgroup of finite index) must also satisfy the second subgroup property (i.e., quotient-powering-invariant subgroup)
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Statement

Suppose $G$ is a group and $H$ is a normal subgroup of finite index in $G$, i.e., $H$ is a normal subgroup of $G$ and the index of $H$ in $G$ is finite. Then, $H$ is a quotient-powering-invariant subgroup of $G$, i.e., if $G$ is powered over a prime number $p$, then so is the quotient group $G/H$.

Related facts

• Finite normal implies quotient-powering-invariant
• Powering-invariant and central implies quotient-powering-invariant
• Normal of finite index implies completely divisibility-closed

Facts used

1. Divisibility is inherited by quotient groups
2. Finite and p-divisible implies p-powered

Proof

Proof using given facts (less explicit)

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Given: A group $G$, a normal subgroup $H$ of $G$ such that the index of $H$ in $G$ is finite (in other words, the quotient group $G/H$ is a finite group). $G$ is powered over a prime $p$.

To prove: $G/H$ is also powered over $p$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$G$ is $p$-divisible.		$G$ is $p$-powered.		By definition, being powered over a prime $p$ means every element has a unique $p^{th}$ root, which implies that every element has a $p^{th}$ root (the definition of $p$-divisible).
2	$G/H$ is $p$-divisible.	Fact (1)	$H$ is normal in $G$.	Step (1)	Step-fact combination direct.
3	$G/H$ is $p$-powered.	Fact (2)	$H$ is normal of finite index, so $G/H$ is a finite group.	Step (2)	Step-fact-given direct.

Hands-on proof

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This is the same as the preceding proof, but shows all the details explicitly without appeal to separate facts.

Given: A group $G$, a normal subgroup $H$ of $G$ such that the index of $H$ in $G$ is finite (in other words, the quotient group $G/H$ is a finite group). A prime number $p$ such that for any $g\in G$, there is a unique $x\in G$ such that $x^p=g$.

To prove: For every $a\in G/H$, there is a unique $b\in G/H$ such that $b^p=a$.

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

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The map $x\mapsto x^p$ in $G$ descends to the corresponding map $x\mapsto x^p$ in the quotient group $G/H$.		$H$ is normal in $G$.		Direct from definition of quotient group structure.
2	The map $x\mapsto x^p$ in $G/H$ is surjective from $G/H$ to itself. In other words, for any $a\in G/H$, there exists $b\in G/H$ such that $b^p=a$.		$G$ is $p$-powered.		For any $a\in G/H$, pick $g\in G$ in that coset (so $\phi (g)=a$). There exists $x\in G$ such that $x^p=g$, since $G$ is $p$-powered. Let $b=\phi (x)$ be the coset of $x$. Then, we have that $b^p=a$.
3	The map $x\mapsto x^p$ in $G/H$ is bijective from $G/H$ to itself. In other words, for any $a\in G/H$, there exists $b\in G/H$ such that $b^p=a$.		$H$ is normal of finite index, so $G/H$ is a finite group.	Step (2)	Follows directly from Step (2) and the observation that any surjective map from a finite set to itself must be bijective.