# Let $G$ be a finite group. If $a = bab$, is it true that $b^{2} = e$

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?

• 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 $n$ – Sorfosh Nov 1 '18 at 3:56
• To find counter examples it's always a good idea to check Quaternion group as a first guess.. – yathish Nov 1 '18 at 4:33

No. For example, in $$Q_8$$ we have $$jij=jk=i$$.
Another counterexample is in $$D_3$$ where we have $$f = R_{120}fR_{120}$$.