Prove that any element x, in a group, that satisfies x^9 = e, and x^11 = e, where e is the group identity, must itself be the identity.

Question

anonymous · Answer

so far I've got $$x^9 = e = x ^{11}$$ and $$x ^{11} = x ^{2}x ^{9} = x ^{2}e= x ^{2}$$

anonymous · Answer

Im just having trouble coming up with a valid last step to show that  x=e

anonymous · Answer

ah it was the inverse I wasn't considering, thanks! Silly last step had me furiously sipping my coffee and wondering what was missing. Much appreciated!

anonymous · Answer

Have you done any topology? I'm looking for someone to have a look at my question.

anonymous · Answer

Not yet unfortunately :(

anonymous · Answer

Alright, thanks for having a look though.

anonymous · Answer

Sorry I made a mistake earlier. This should work though
$e = e^{-5} = (x^{2})^{-5} = x^{-10}$
$e = e\cdot e = x^{-10} \cdot x^{11} = x$