TITLE: Solve $x^7=e$ in a group
QUESTION [6 upvotes]: Let $(G,\cdot)$ be a group having the property that $\exists a \in G$ such that $\forall x \in G$, $ax=x^4a$. Solve the equation $x^7=e$.
I started by observing that for $x=a$ we have that $a^2=a^5$, so $a^3=e$. Then I tried to left multiply the relation in the hypothesis by $x^3$, but it didn't help and now I am stuck.
EDIT: This is what the answer key says "take $a=x^7$ to get that $x^8=x^{11} \implies x^3=e$. Hence, $a=x^7=(x^3)^2 x=x \implies x=a$, which satisfies $x^7=e$".
To me, this seems blatantly wrong. Firstly, I don't think we can set $a$ to be equal to anything because we are only told that it exists, nothing more. Secondly, $a^7=a$ since $a^3=e$.
Could it be possible that this problem is wrong and the equation in fact has no solution?

REPLY [5 votes]: As shown in the comments, $x^{63}=e,\,\forall x\in G$. Hence, for every $x$ in $G$, $(x^9)^7=e$. Conversely, if $x^7=e$, then $x=(x^4)^9$. So the solutions to the equation $x^7=e$ form the set $\left\{x^9\mid x\in G\right\}$.

For the sake of completeness, let's show $x^{63}=e,\,\forall x\in G$ here. Since for every $x$ in $G$, $ax^4a^{-1}=x^{16}$, we see $x^{16}=a^2xa^{-2}$, and hence $x^{64}=a^3xa^{-3}=x$, thus $x^{63}=e$.


Hope this helps.