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

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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:

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

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.