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

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

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

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

6 Answers 6

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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.

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

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$(ab)^n=e$, 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: $ab=a(ba)a^{-1}$

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

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

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

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