Equivalence of definitions of nilpotent group that is torsion-free for a set of primes

For an arbitrary (not necessarily nilpotent) group and a prime
Suppose $$G$$ is a group and $$p$$ is a prime number. We have the implications (1) implies (2) implies (3) implies (4) implies (5) implies (6):


 * 1) $$G$$ is a $$p$$-powering-injective group, i.e., $$x \mapsto x^p$$ is injective.
 * 2) $$G$$ is a $$p$$-torsion-free group.
 * 3) There exists an element $$g \in G$$ such that the equation $$x^p = g$$ has a unique solution for $$x \in G$$.
 * 4) The center $$Z(G)$$ is a $$p$$-torsion-free group.
 * 5) Each of the successive quotients $$Z^{i+1}(G)/Z^i(G)$$ in the upper central series of $$G$$ is a $$p$$-torsion-free group.
 * 6) All quotients of the form $$Z^i(G)/Z^j(G)$$ for $$i > j$$ are $$p$$-powering-injective groups, i.e., $$x \mapsto x^p$$ is injective in each such quotient group.

For a nilpotent group and a prime
Suppose $$G$$ is a nilpotent group and $$p$$ is a prime number. The following are equivalent:


 * 1) $$G$$ is a $$p$$-powering-injective group, i.e., $$x \mapsto x^p$$ is injective.
 * 2) $$G$$ is a $$p$$-torsion-free group.
 * 3) There exists an element $$g \in G$$ such that the equation $$x^p = g$$ has a unique solution for $$x \in G$$.
 * 4) The center $$Z(G)$$ is a $$p$$-torsion-free group.
 * 5) Each of the successive quotients $$Z^{i+1}(G)/Z^i(G)$$ in the upper central series of $$G$$ is a $$p$$-torsion-free group.
 * 6) All quotients of the form $$Z^i(G)/Z^j(G)$$ for $$i > j$$ are $$p$$-powering-injective groups, i.e., $$x \mapsto x^p$$ is injective in each such quotient group.

For an arbitrary (not necessarily nilpotent) group and a set of primes
Suppose $$G$$ is a group and $$\pi$$ is a set of prime numbers. We have the implications (1) implies (2) implies (3) implies (4) implies (5) implies (6):


 * 1) $$G$$ is a $$\pi$$-powering-injective group, i.e., $$x \mapsto x^p$$ is injective and each $$p \in \pi$$.
 * 2) $$G$$ is a $$\pi$$-torsion-free group.
 * 3) For each $$p \in \pi$$, there exists an element $$g \in G$$ (possibly dependent on $$p$$) such that the equation $$x^p = g$$ has a unique solution for $$x \in G$$.
 * 4) The center is a $$\pi$$-torsion-free group.
 * 5) Each of the successive quotients $$Z^{i+1}(G)/Z^i(G)$$ in the upper central series of $$G$$ is a $$\pi$$-torsion-free group.
 * 6) All quotients of the form $$Z^i(G)/Z^j(G)$$ for $$i > j$$ are $$\pi$$-powering-injective groups, i.e., $$x \mapsto x^p$$ is injective in each such quotient group and each $$p \in \pi$$.

For a nilpotent group and a set of primes
Suppose $$G$$ is a nilpotent group and $$\pi$$ is a set of prime numbers. The following are equivalent:


 * 1) $$G$$ is a $$\pi$$-powering-injective group, i.e., $$x \mapsto x^p$$ is injective and each $$p \in \pi$$.
 * 2) $$G$$ is a $$\pi$$-torsion-free group.
 * 3) For each $$p \in \pi$$, there exists an element $$g \in G$$ (possibly dependent on $$p$$) such that the equation $$x^p = g$$ has a unique solution for $$x \in G$$.
 * 4) The center is a $$\pi$$-torsion-free group.
 * 5) Each of the successive quotients $$Z^{i+1}(G)/Z^i(G)$$ in the upper central series of $$G$$ is a $$\pi$$-torsion-free group.
 * 6) All quotients of the form $$Z^i(G)/Z^j(G)$$ for $$i > j$$ are $$\pi$$-powering-injective groups, i.e., $$x \mapsto x^p$$ is injective in each such quotient group and each $$p \in \pi$$.

Dual
The dual fact to this is dual::equivalence of definitions of nilpotent group that is divisible for a set of primes.

The duality is as follows:

Facts used

 * 1) uses::Powering-injectivity is inherited by central extensions

Proof
We first prove the directions of implication for an arbitrary group and a single prime, then complete the equivalences by showing (6) implies (1) for the nilpotent case. The version for a prime set follows.

(1) implies (2)
If $$x \mapsto x^p$$ is injective, then since the identity element also has itself as a $$p^{th}$$ root, it cannot have any other $$p^{th}$$ roots. Thus, $$G$$ is $$p$$-torsion-free.

(2) implies (3)
Take the element $$g$$ to be the identity element of $$G$$.

(3) implies (4)
Given: A group $$G$$, a prime number $$p$$, an element $$g \in G$$ such that $$x^p = g$$ has a unique solution $$x \in G$$. An element $$z \in Z(G)$$ (where $$Z(G)$$ is the center of $$G$$) such that $$z^p $$ is the identity element of $$Z(G)$$ and hence also of $$G$$.

To prove: $$z$$ is the identity element of $$G$$ (and hence also of $$Z(G)$$).

Proof:

(4) implies (5)
This is the trickiest part and in some sense the meat of the proof.

We will prove the claim by induction. Explicitly, the claim is that for every nonnegative integer $$i$$, the quotient group $$Z^{i+1}(G)/Z^i(G)$$ is $$p$$-torsion-free.

Base case for induction: In case $$i = 0$$, the quotient becomes $$Z(G)$$, which is $$p$$-torsion-free by assumption.

Inductive step:

Inductive hypothesis: The quotient group $$Z^i(G)/Z^{i-1}(G)$$ is $$p$$-torsion-free.

Goal of inductive step: The quotient group $$Z^{i+1}(G)/Z^i(G)$$ is $$p$$-torsion-free.

We now frame our inductive step explicitly:

Given: An element $$x \in Z^{i+1}(G)$$ such that $$x^p \in Z^i(G)$$

To prove: $$x \in Z^i(G)$$

Proof:

(5) implies (6)
This relies on Fact (1), and the observation that for abelian groups, being powering-injective is the same as being torsion-free. We can write the proof formally by inducting on the magnitude of the difference $$i - j$$.

(6) implies (1): valid only in the nilpotent case
If $$G$$ has class $$c$$, set $$i = c, j = 0$$ to get the result.