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Powering-injectivity is central extension-closed

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-powering-injective, i.e., x \mapsto x^p is injective from H to itself.
  • The quotient group G/H is p-powering-injective, i.e., x \mapsto x^p is injective from G/H to itself.

Then, the whole group G is p-powering-injective.

Related facts

Proof

Given: A group G, a prime number p. A central subgroup H of G such that in both H and G/H viewed separately, the map x \mapsto x^p is injective. Two elements a,b \in G with a^p = b^p.

To prove: a = b.

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

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 \pi(a) = \pi(b). a^p = b^p, and the map x \mapsto x^p is injective in G/H. We have a^p = b^p, hence \pi(a^p) = \pi(b^p). Thus, \pi(a)^p = \pi(b)^p, so \pi(a),\pi(b) are elements of G/H with the same p^{th} power. Thus, they must be equal.
2 The element u = ab^{-1} is an element of H. Step (1) \pi(u) = \pi(b^{-1}) = \pi(a)\pi(b)^{-1} is the identity element of G/H, so u \in H.
3 u^p is the identity element of H. a^p = b^p, H is central in G. Step (2) We have a = ub. Thus, a^p = (ub)^p. By the centrality of H, we get math>a^p = u^pb^p</math>. We also have a^p = b^p. This gives b^p = u^pb^p, so u^p is the identity element of G, and hence also of the subgroup H.
4 u is the identity element of H. x \mapsto x^p is injective in H. Step (3) Step-given direct.
5 a = b. Steps (2), (4) Step-combination direct.