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Let $G$ be a finite group and let $a,b \in G$. If $a = bab$, is it true that $b^{2} = e$. If not, find a counterexample.

It is clear that if $a = bab$ and $b^{2} = e$ are both true, then $ab = ba$. However, there exist groups (namely non-Abelian ones) with elements such that $ab \neq ba$. However, I am having trouble finding a non-Abelian group with elements such that $a = bab$, but $ab \neq ba$. How does one solve this problem?

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  • $\begingroup$ I am not sure how helpful it is but you can say $a=bab$ and the substitute $a$ to get $a=bbabb$ Etc. so $a=b^nab^n$ for any $n$ $\endgroup$
    – Sorfosh
    Commented Nov 1, 2018 at 3:56
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    $\begingroup$ To find counter examples it's always a good idea to check Quaternion group as a first guess.. $\endgroup$
    – yathish
    Commented Nov 1, 2018 at 4:33

2 Answers 2

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No. For example, in $Q_8$ we have $jij=jk=i$.

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Another counterexample is in $D_3$ where we have $f = R_{120}fR_{120}$.

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