# If $a,b \in$ group $G$ such that $a^2=e, a*b^4*a=b^7$, prove that $b^{33}=e$

$e$ is the identity of the group.

My understanding: To prove that $b^{33}=e$ is the same as proving $b^{34}=b$

Now, $a * b^4 * a=b^7$

$\Rightarrow b^4= a*b^7*a=(a*b*a)^7$

For any integer $n$ we have $$(ab^4a)^n=b^{7n}$$ or $$ab^{4n}a=b^{7n}$$ For $n=4$, $$ab^{16}a=b^{28}$$ and for $n=7$, $$ab^{28}a=b^{49}$$ Therefore $$b^{16}=b^{49}$$

EDIT: With same technique, it can be shown that if $a^2=e$ and $ab^r=b^sa$ then $b^{r^2}=b^{s^2}$.

• Another solution: Note that $a$ and $b^{11}$ commute; then take $n=11$ above to get $b^{44}=b^{77}$. – Aravind Jul 14 '15 at 9:27
• @ajotatxe Nice ... Vote up! – johannesvalks Jul 14 '15 at 9:29
• Very nice, +1 Then one could note that in the dihedral group of order $66$, generated by an element $c$ of order $33$, and by an element $a$ of order $2$, one has $c a c = c^{-1}$, and for $b = c^{3}$, one has $a b^4 a = a c^{12} a = c^{-12} = c^{21} = (c^{3})^{7}= b^7$. So $b^{33} = 1$ is the best we can squeeze out of the relations. – Andreas Caranti Jul 14 '15 at 17:09

Note that \begin{align} (aba^{-1})^4&=b^7\text{, and }\\ b^4&=(aba^{-1})^7\nonumber\\ (aba^{-1})^7&=b^4. \text{So 'dividing' the first equality by the last,}\\ (aba^{-1})^{-3}&=b^{3}. \text{ Multiplying it with the first equality,}\\ aba^{-1}&=b^{10}\\ (aba^{-1})^4&=b^{40}=b^7\\ \text{So }b^{33}&=e \end{align}

$$\textbf{Different approach:}$$

$$ab^4a=b^7$$ and $$ab^7a=b^4$$. So $$b^{11}=b^4b^7=ab^{11}a$$ Hence, $$(b^4)^{11}=(ab^7a)^{11}=ab^{77}a=(b^{11})^4=(ab^{11}a)^4=ab^{44}a$$. Hence, $$b^{44}=b^{77}$$