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

This article gives a proof/explanation of the equivalence of multiple definitions for the term nilpotent group that is torsion-free for a set of primes
View a complete list of pages giving proofs of equivalence of definitions

## Statement

### 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 $p \in \pi$.

## Related facts

### Dual

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

The duality is as follows:

Concept of the torsion-free side Concept on the divisible side
torsion-free divisible
center abelianization
inner automorphism group derived subgroup
upper central series lower central series

## Facts used

1. 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 can be deduced easily, so we do not include it explicitly.

### (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:

Step no. Assertion/construction Given data used Previous steps used Explanation
1 $(xz)^p = x^pz^p$ $z$ is in the center of $G$
2 $(xz)^p = x^p = g$ $z^p$ is the identity element of $G$. Step (1)
3 $xz = x$. $g$ has a unique $p^{th}$ root. Step (2) Step-given direct.
4 $z$ is the identity element of $G$. Step (3) Step-direct

### (4) implies (5)

The proof method used in this article is discussed in the survey article upward induction on upper central series.|See a list of facts whose proof uses this method

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:

Step no. Assertion/construction Facts used Given data/inductive hypothesis used Previous steps used Explanation
1 For every $y \in G$, $[x^p,y] \in Z^{i-1}(G)$ $x^p \in Z^i(G)$. Also, we are only using the central series nature, not the upper central series nature. By definition, $[Z^i(G),G] \le Z^{i-1}(G)$, so this follows.
2 $[x^p,y] \equiv [x,y]^p \pmod{Z^{i-1}(G)}$ for all $y \in G$. Formula for commutator of element and product of two elements Consider the commutator map as a map $G/Z^{i-1}(G) \times Z^{i+1}(G)/Z^{i-1}(G) \to Z^i(G)/Z^{i-1}(G)$. By the formula for the commutator of the product, we obtain that the commutator map is a bihomomorphism, i.e., it is a homomorphism in each coordinate holding the other constant. This gives the equality.
3 $[x,y]^p \in Z^{i-1}(G)$ for all $y \in G$. Steps (1), (2) Step-combination direct.
4 $[x,y] \in Z^i(G)$ for all $y \in G$. $x \in Z^{i+1}(G)$. Also, we are only using the central series nature, not the upper central series nature. By definition, $[Z^{i+1}(G),G] \le Z^i(G)$, so this follows.
5 $[x,y] \in Z^{i-1}(G)$ for all $y \in G$. Inductive hypothesis about $Z^i(G)/Z^{i-1}(G)$ being $p$-torsion-free. Steps (3), (4) Consider the image of $[x,y]$ mod $Z^{i-1}(G)$, viewed as an element of $Z^i(G)/Z^{i-1}(G)$. This is an element of $Z^i(G)/Z^{i-1}(G)$ whose $p^{th}$ power is the identity element. By the inductive hypothesis, $Z^i(G)/Z^{i-1}(G)$ is $p$-torsion-free, so the element is the identity element, and therefore, $[x,y] \in Z^{i-1}(G)$.
6 $x \in Z^i(G)$. This is the step where we use that we are dealing with the upper central series Step (5) We can interpret Step (5) as saying that the image of $x$ in $G/Z^{i-1}(G)$ is central in $G/Z^{i-1}(G)$. The center of $G/Z^{i-1}(G)$ is $Z^i(G)/Z^{i-1}(G)$, so the image of $x$ is in $Z^i(G)/Z^{i-1}(G)$. Thus, $x \in Z^i(G)$.

### (5) implies (6)

This proof uses the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).

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.