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Let $G$ be a group and $a,b \in G$. Prove that if $a^{2}=e$ and $ab^{4}a=b^{7}$, then $b^{33}=e$, where $e$ is the identity of a group $G$.

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More generally, $a^2=e$ and $a b^p a = b^q$ implies $b^{p^2}=b^{q^2}$. You can use anon's proof for this. – Martin Brandenburg Oct 3 '13 at 12:58
up vote 15 down vote accepted

So $\rm ab^4a=b^7$. Apply $\rm a$ on left and right to get $\rm ab^7a=b^4$. Then, using symmetry in calculation,

$$\rm ab^{4\times7}a=\begin{cases}(ab^4a)^7=(b^7)^7=b^{49} \\[3pt] \rm(ab^7a)^4=(b^4)^4=b^{16} \end{cases}$$

and so $\rm b^{49}=b^{16}\iff e=b^{49-16}=b^{33}$.

Note that since $\rm a^2=e\iff a=a^{-1}$, the map $\rm\Phi_a: x\mapsto axa=axa^{-1}$ is conjugation, which is a special type of automorphism. In particular $\rm\Phi_a(xy)=\Phi_a(x)\Phi_a(y)$ and thus $\rm\Phi_a(x^{n})=\Phi_a(x)^n$ for any elements $\rm x,y$ and integer ${\rm n}\in{\bf Z}$. We employed the latter exponential distributivity above.

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thank you so much – nameless Mar 22 '13 at 8:23
As an exercise, you could generalize this for $a^2 = e$, $a b^n a = b^m$. – Mikko Korhonen Mar 22 '13 at 13:34

$$b^{49} = (b^7)^7 = (ab^4a)^7 = a (b^7)^4 a = a (a b^4 a)^4 a = b^{16}.$$

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Since $a$ is its own inverse, $b^4 = ab^7a = ab^8b^{-1}a = ab^4a ab^4a ab^{-1}a = b^{14} ab^{-1}a$. Therefore, $ab^{-1}a = b^{-10}$. Inverting both, $aba = b^{10}$. Now, $b^7 = ab^4a = abaabaabaaba = b^{40}$ gives the result.

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Nice solution.! – nameless Mar 22 '13 at 6:17

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