View source for Powering-invariant over quotient-powering-invariant implies powering-invariant

==Statement==

Suppose <math>G</math> is a [[group]] and <math>H,K</math> are subgroups of <math>G</matH> such that <matH>H \le K \le G</math>, <math>H</matH> is normal in <math>G</math>, and the following conditions hold:

* <math>H</math> is a [[quotient-powering-invariant subgroup]] of <math>G</math>.
* <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==

* [[Powering-invariance is not quotient-transitive]]
* [[Quotient-powering-invariance is quotient-transitive]]

==Proof==

'''Given''': <math>G</math> is a group and <math>H,K</math> are subgroups of <math>G</matH> such that <matH>H \le K \le G</math>, <math>H</matH> is normal in <math>G</math>, and the following conditions hold:

* <math>H</math> is a quotient-powering-invariant subgroup of <math>G</math>.
* <math>K/H</math> is a powering-invariant subgroup of <math>G/H</math>.

<math>p</math> is a prime number such that <math>G</math> is powered over <math>p</math>. An element <math>g \in K</math>.

'''To prove''': There exists <math>x \in K</math> such that <math>x^p = g</math>.

'''Proof''': Let <math>\varphi:G \to G/H</math> be the quotient map.

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