Ambivalent and nilpotent implies 2-group

Statement

Suppose $G$ is a group that is both an ambivalent group and a nilpotent group. Then, $G$ is a 2-group, i.e., the order of every element of $G$ is a power of 2. In fact, $G$ is a group of finite exponent and the log of the exponent to base 2 is at most the nilpotency class of $G$ .

Applications

Rational and nilpotent implies 2-group

Facts used

Proof

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).

We induct on the nilpotency class $c$ .

Base case for induction: An ambivalent nilpotent group of nilpotency class $c=0$ is a 2-group of exponent dividing $2^{0}=1$ .

Proof: This is direct since the only such group is the trivial group.

Inductive step:

Hypothesis: Any ambivalent nilpotent group of nilpotency class $c-1$ is a 2-group of exponent dividing $2^{c-1}$ .

Goal of inductive step: Prove the following:

Given: An ambivalent nilpotent group $G$ of nilpotency class $c$ .

To prove: $G$ is a 2-group of exponent dividing $2^{c}$ .

Proof:

No.	Assertion/construction	Facts used	Given data/inductive assumption used	Previous steps used	Explanation
1	The center $Z(G)$ is an elementary abelian 2-group.	Fact (1)	$G$ is ambivalent		direct
2	The quotient $G/Z(G)$ is nilpotent of class $c-1$ .		$G$ is nilpotent of class $c$ .		definition of nilpotency class.
3	$G/Z(G)$ is a 2-group of exponent dividing $2^{c-1}$ .		inductive assumption	Step (2)	step-assumption-combination-direct.
4	$G$ is a 2-group of exponent dividing $2^{c}$ .			Steps (1), (3)	For any $g\in G$ , Step (3) tells us that $g^{2^{c-1}}\in Z(G)$ . Step (1) tells us that this element squares to the identity, so $(g^{2^{c-1}})^{2}=g^{2^{c}}$ is the identity element.