# Endomorphism kernel 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., endomorphism kernel) 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 an endomorphism kernel in $G$, i.e., $H$ is a normal subgroup of $G$ and there is a subgroup $K$ of $G$ such that $G/H \cong K$. Then, $H$ is a quotient-powering-invariant subgroup of $G$, i.e., for any prime number $p$ such that $G$ is $p$-powered, so is the quotient group $G/H$.

## Proof

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Given: A group $G$, a normal subgroup $H$ of $G$ such that $G/H$ is isomorphic to a subgroup $K$ of $G$. $G$ is powered over a prime number $p$.

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

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let $\varphi:G \to K$ be the composite of the quotient map $G \to G/H$ and the isomorphism from $G/H$ to $K$. $H$ is normal in $G$ and $G/H \cong K$.
2 For $g \in K$, there exists a unique $x \in G$ such that $x^p = g$. $G$ is powered over $p$.
3 The $x$ obtained in Step (2) is actually an element of $K$. In other words, for $g \in K$, there exists a unique $x \in K$ such that $x^p = g$. So, $K$ is powered over $p$. Steps (1), (2) Suppose $u \in \varphi^{-1}(\{ g \})$ (note that such a $u$ exists because the image of $\varphi$ is $K$). Then, since $u \in G$, there exists $v \in G$ such that $v^p = u$. Thus, $(\varphi(v))^p = \varphi(u) = g$, with $\varphi(v) \in K$. By the uniqueness of $x$ from Step (2), $\varphi(v) = x$. Thus, $x \in K$.
4 $G/H$ is powered over $p$. $G/H \cong K$. Steps (1), (3) Step (1) says that $G/H \cong K$. Step (3) says that $K$ is powered over $p$. Thus, $G/H$ is powered over $p$.