# Prove that if $(ab)^2 = e$, then $(ba)^2 = e$ [closed]

Let $G$ be a non-abelian group, and let $a,b \in G$ be such that

$$(ab)^2 = e.$$

Prove that $(ba)^2 = e$.

• Did you have any thoughts of your own on the matter? Aug 10, 2013 at 16:53
• (ab)^-1=b^-1 a^-1=ab . b(b^-1 a^-1) a = b(ab)a=e Oct 30, 2016 at 12:52

Hint:

If $(ab)^{2} = abab = e$, right multiply both sides by $a$. Then we have $a(baba) = a$. Now use the uniqueness of the identity.

Hint: Consider $a^{-1}ea$ and see what gives.

$(ab)^n=e$, then $(ba)^n=e$. Here are two proofs:

1. In fact, $(ab)^n=e$, so $$b(ab)^n=b$$ associative law $$(ba)^nb=b$$ we get that $(ba)^n=e$.

2. $ab$ and $ba$ conjugacy elements: $ab=a(ba)a^{-1}$

$$abab = e$$ $$a^{-1}abab = a^{-1}$$ $$bab= a^{-1}$$ $$baba = a^{-1}a$$ $$baba = e$$

Given: $(ab)^2 = e$.

Knowing that $$(ab)^{2} = abab = e$$ then using right multiplication by $a$, we have $$(abab)a = ea = a \iff a(baba) = a$$

But the identity $e$ in any group is unique. Hence, $$\;a(baba) = {\bf a}(ba)^2 = {\bf a} \implies (ba)^2 = e$$

That is, we have both $(ab)^2 = (ba)^2 = e$.

Useful fact to prove: for all $a,b\in G$ [ $(bab^{-1})^n=ba^nb^{-1}$]