prove the sub ring forms a group under multiplication operation

Question

Let $(R,+,\cdot)$ be a unitary ring such that $a^2\neq0$ for all non-zero $a\in R$, and let $H=\{b \in R\ \mid\ $$b^2=1\}.$ If the expression $ab-ba=bab-a$ holds true $\forall a \in R, b\in H$, then $(H,\cdot)$ forms a group.

Here's what I've got so far: Take any $a,b\in H$, hence $(ab-ba)^2=-(bab-a)^2$. Then $2(ab-ba)^2=0$, so that $2ab=2ba$.

From here I got stuck (I'm trying to prove $(ab)^2=1$). How do I proceed?

the question states only unitary ring. i guess we have to prove it is commutative — user1259172, Jun 02 '24 at 11:44
Is the part "we have $ab-ba=bab-a$" from you, or was it in the original problem formulation? Do you have a link to the original question? — Dietrich Burde, Jun 02 '24 at 12:12
@user1259172 A group whose elements square to 1 is necessarily commutative. Searching for "ab-ba=bab-a" via Approach0 gave the following result, see the penultimate post. — Amateur_Algebraist, Jun 02 '24 at 18:31

score 1 · Answer 1 · answered Jun 05 '24 at 19:23

In order to show that $H$ defines a group (with respect to the multiplicative operator on $R$), it's enough to show that it satisfies the four axioms of a group:

Well-Defined Binary Operator. Let $b\neq1$. Given $a,b\in H$, then: $$-=-\ \Rightarrow\ abbab-babab=babbab-abab\ \Rightarrow\ b-1=(b-1)abab.$$ Since $b\neq1$ by assumption, and since $(abab)(baba)=1$ (i.e. $abab$ is a unit and therefore cannot be a zero divisor), it must follow that $(ab)^2=abab=1$ (that is, $ab\in H$). And therefore the multiplicative binary operator will behave in a well-defined manner when restricted from $R$ onto $H$.
Identity. Observe that $1^2=1$. Hence, $1\in H$.
Associativity. Inherited from ring structure of $R$.
Inverse. This'll hold true noting that $a^2=1$ implies $a^{-1}$ will exist and is $=a$.

From this, we conclude $H$ defines a group under the inherited multiplicative binary operator from ring $R$.

prove the sub ring forms a group under multiplication operation

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