Divisibility is central extension-closed

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This article gives the statement, and possibly proof, of a group property (i.e., divisible group for a set of primes) satisfying a group metaproperty (i.e., central extension-closed group property)
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Statement

Suppose G is a group and H is a central subgroup of G. Suppose p is a prime number such that:

  • H is p-divisible.
  • The quotient group G/H is p-divisible.

Then, the whole group G is p-divisible.

Related facts

Dual fact

The dual fact is that powering-injectivity is inherited by central extensions.

Other related facts

Proof

Proof idea

The idea is that of successive approximation. We first obtain a p^{th} root in the quotient group, then pick a representative. We then pick a representative, and measure the extent to which its p^{th} power misses the mark. Then, we take a p^{th} root of that, and use that as the "first-order correction" to our original choice of representative.

The subgroup being central is crucial in making sure that the product of the p^{th} powers of the original representative and the correction is the p^{th} power of the product.

Proof details

Given: A group G, a central subgroup H of G, a prime p such that both H and G/H are p-divisible. An element g \in G.

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

Proof: Suppose \pi:G \to G/H is the quotient map.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation Commentary
1 Let a = \pi(g), so a \in G/H. moved to quotient group
2 There exists b \in G/H such that b^p = a. G/H is p-divisible. Step (1) direct took root in quotient group
3 Suppose y \in \pi^{-1}(b), i.e., y \in G is such that \pi(y) = b. Then, u = y^{-p}g is an element of H. Steps (1), (2) We have that \pi(y^{-p}g) = b^{-p}a which is the identity element of G/H. Thus, y^{-p}g \in H. picked representative, this is our "first guess" y, and measured how far it is, using u.
4 There exists v \in H such that v^p = u. H is p-divisible. Step (3) direct picked v as the "correction" for our first guess.
5 The element x = yv works, i.e., x^p = g. H is central. Steps (3), (4) We have (yv)^p = y^pv^p because v \in H and H is central. Thus, x^p = (yv)^p = y^pv^p = y^pu = y^py^{-p}g = g, as desired. multiplied the first guess and the correction to get a correct guess.