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

 * 1) uses::Center of ambivalent group is elementary abelian 2-group
 * 2) uses::Ambivalence is quotient-closed

Proof
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: