Difference between revisions of "Powering-invariant over quotient-powering-invariant implies powering-invariant"

Latest revision as of 16:00, 19 December 2014

Suppose $G$ is a group and $H,K$ are subgroups of $G$ such that $H \le K \le G$ , $H$ is normal in $G$ , and the following conditions hold:

Then, $K$ is a powering-invariant subgroup of $G$ .

Given: $G$ is a group and $H,K$ are subgroups of $G$ such that $H \le K \le G$ , $H$ is normal in $G$ , and the following conditions hold:

$p$ is a prime number such that $G$ is powered over $p$ . An element $g \in K$ .

To prove: There exists $x \in K$ such that $x^p = g$ .

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

Step no.	Assertion/construction	Given data used	Previous steps used	Explanation
1	There exists $x \in G$ such that $x^p = g$ .	$G$ is $p$ -powered.		Given-direct.
2	$G/H$ is $p$ -powered.	$H$ is quotient-powering-invariant in $G$ , $G$ is $p$ -powered.		Given-direct.
3	$(\varphi(x))^p = \varphi(g)$ and $\varphi(x)$ is the unique $p^{th}$ root of $\varphi(g)$ in $G/H$ .		Steps (1), (2)	Applying the homomorphism $\varphi$ to Step (1) gives that $(\varphi(x))^p = \varphi(g)$ . Since, by Step (2), $G/H$ is $p$ -powered, $\varphi(x)$ must be the unique $p^{th}$ root.
4	$\varphi(x) \in K/H$ .	$K/H$ is powering-invariant in $G/H$ .	Steps (2), (3)	By Step (2), $G/H$ is $p$ -powered, hence $K/H$ is $p$ -powered from the given information. Thus, the unique $p^{th}$ root of $\varphi(g)$ in $G/H$ , which equals $\varphi(x)$ ,(from Step (3)) must be in $K/H$ .
5	$x \in K$ .	$H \le K$	Step (4)	Since $\varphi(x) \in K/H$ , $x \in \varphi^{-1}(K/H)$ , which is $K$ since $H \le K$ .

@@ Line 6: / Line 6: @@
 * <math>K/H</math> is a [[powering-invariant subgroup]] of <math>G/H</math>.
+Then, <math>K</math> is a powering-invariant subgroup of <math>G</math>.
 ==Related facts==