Proof help: a² = identity -> monoid must be a commutative group

Question

We have a monoid with the identiy $e$, and we know that any element of the monoid squared is the identity. I have to prove now that this is a commutative group.

May somebody help me or give me a hint how I can show that this must be true?

Answer 1

To prove that $a^2 = e$ for all $a$ implies $ab = ba$:

$(ab)(ab) = e$

$a(ab)(ab)b = a*e*b$

$(aa)(ba)(bb) = ab$

$e*ba*e = ab$

$ba = ab$.

or

$ab = k$

$a=abb = kb$

$e = aa = kba$

$k = ke = kkba = ba$

So $ab = k = ba$

Answer 2

$a^2=e$ implies that every $a$ has an inverse, itself. Therefore, this monoid is in fact a group. It is well known that $a^{2}=e$ in a group implies that it is commutative.

Answer 3

To show it's a commutative group, you need to show two things:

• Existence of inverses
• Commutativity

For inverses, take any element $a$; $a^2 = e$, so $a$ is its own inverse.

For commutativity, you need to show that for any elements $a,b$, $ab = ba$. Since $a^{-1}=a$ and $b^{-1}=b$, $$ab(ba)^{-1} = aba^{-1}b^{-1}=abab = (ab)^2 = e$$ Thus (multiplying both sides by $ba$ on the right) $ab = ba$