Finite normal 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., finite normal subgroup) 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 finite normal subgroup of $G$. Then, $H$ is a quotient-powering-invariant subgroup of $G$, i.e., for any prime number $p$ such that $G$ is powered over $p$, so is the quotient group $G/H$.

Facts used

1. Left cosets are in bijection via left multiplication

Proof

Proof idea

This is relatively straightforward, and involves using the fact that an injective map between subsets of equal cardinality (here, the subsets are cosets of $H$, and the map is the $p^{th}$ power map) must be surjective, and hence the inverse to it is defined. Note that the key way the proof fails for infinite group is that it is possible for the powering map between cosets of an infinite normal subgroup to be such that a given coset is a disjoint union of the images under the $p^{th}$ power map of more than one coset, even though all the maps are injective.

For instance, for the case $G=\mathbb {Q} ,H=\mathbb {Z} ,p=2$, the identity coset is not covered as the set of squares of any single coset, but is the union of the set of squares of the cosets $\mathbb {Z}$ and ${\frac {1}{2}}+\mathbb {Z}$.

Proof details

Given: A group $G$, a finite normal subgroup $H$ of $G$. A prime number $p$ such that for any $g\in G$, there exists a unique $x\in G$ such that $x^{p}=g$.

To prove: For any $a\in G/H$, there exists $b\in G/H$ such that $b^{p}=a$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The map $x\mapsto x^{p}$ in $G$ sends each coset of $H$ to within a coset of $H$, namely, its $p^{th}$ power in the quotient group structure on $G/H$.		$H$ is normal in $G$.		Direct from definition of quotient group structure.
2	The restriction of the map $x\mapsto x^{p}$ to any coset is injective from that coset to the target coset (namely, the coset in which the image lands).		$G$ is powered over $p$, i.e., every element has a unique $p^{th}$ root.	Step (1)	By the fact that $G$ is powered over $p$, two different elements cannot have the same $p^{th}$ power, so the mapping is injective.
3	The restriction of the map $x\mapsto x^{p}$ to any coset is bijective from that coset to the target coset (namely, the coset in which the image lands).	Fact (1)	$H$ is finite	Step (2)	Since $H$ is finite, all its cosets are finite and of the same finite size via Fact (1). Thus, the injective map of Step (2) is bijective.
4	Every coset of $H$ in $G$ is contained in the image of at least one coset under the map $x\mapsto x^{p}$.		$G$ is powered over $p$, i.e., every element has a $p^{th}$ root.		Pick any element in the coset, find a $p^{th}$ root, and take the coset of that.
5	Under the map $x\mapsto x^{p}$, every coset $a$ of $H$ in $G$ is the full image of exactly one coset $b$ of $H$ in $G$, and is not contained in the image of any other coset.		$G$ is powered over $p$, i.e., every element has a unique $p^{th}$ root.	Steps (3), (4)	By Step (4), we can find a coset $b$ whose image under the $p^{th}$ power map contains some elements of $a$. By Step (3), this forces that the $p^{th}$ power map sends the elements of the coset $b$ bijectively to $a$, i.e., every element of the coset $a$ has a $p^{th}$ root in the coset $b$. By the uniqueness of $p^{th}$ roots, this means that there can be no element outside the coset $b$ whose $p^{th}$ power lies in $a$.
6	For any $a\in G/H$, there is a unique $b\in G/H$ such that $b^{p}=a$.			Step (5)	Just reinterpret the preceding step in the new language.