Powering-invariant over quotient-powering-invariant implies powering-invariant

From Groupprops
Revision as of 02:28, 31 March 2013 by Vipul (talk | contribs) (Created page with "==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 <ma...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

Statement

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:

Related facts

Proof

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:

  • H is a quotient-powering-invariant subgroup of G.
  • K/H is a powering-invariant subgroup of G/H.

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 Facts used 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.