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In a group $G$:

If $x^2 a x=a^{-1}$, then $a$ has a cube root. (Hint: Show that $xax$ is a cube root of $a^{-1}$.)

So essentially $\exists y\in G:a=y^3$. The hint probably confused me more than anything. How should I approach this? I am looking for hints, not complete solutions. Thanks.

Patrick Shambayati
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    Just follow the hint: compute $(xax)^3$ using what you know about $x^2ax$. You should find that it simplifies substantially... – Steven Stadnicki Aug 21 '14 at 15:06
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    Did you take the time to digest the answers to your previous question on this very topic? If I were trying to master the subject, this is what I would do before firing new questions. – Did Aug 21 '14 at 15:56
  • Using the "technique" I learned in that previous question, I was able to tackle many more problems. But I am genuinely stuck on this one. On the last problem, there was no step of showing that $a^{-1}$ has a root and then implying that $a$ has root, so this one feels different, I guess – Patrick Shambayati Aug 21 '14 at 15:57

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$(xax)(xax)=xax^2ax=xaa^{-1}=x$, so

$(xax)^3=(xax)(xax)(xax)=x^2ax=a^{-1}$

Nicky Hekster
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Okay, so over here I found a nice solution:

$$x^2ax=a^{-1}$$ $$xax=x^{-1}a^{-1}$$ $$xaxax=x^{-1}a^{-1}ax=e$$ $$xaxaxa=a=(xa)^3$$

This doesn't seem to rely directly on that hint in the problem, but it looks like it works.

Also, you may want to mark my question as a duplicate since I discovered a very similar question in that link.

Thanks

Community
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Patrick Shambayati
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To expand on their hint a bit: $(xax)^3$ $= (xax)\cdot(xax)\cdot(xax)$ $= xax^2axxax$ $=xa(x^2ax)xax$ $= xaa^{-1}xax$ $=\ldots$?

Of course, once you have this, you'll have a cube root for $a^{-1}$; you need to show why this implies that $a$ itself has a cube root. You should be able to use what you know about cube roots in $\mathbb{R}$ as a model for this; if $y$ satisfies $y^3=z$, can you find a number $w$ that satisfies $w^3=z^{-1}$? Once you know what you're looking for, you can try and prove it in the group theory context; this should be straightforward.

Steven Stadnicki
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