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Let $(G,.)$ be a group where there exists an element $g \in G$ such that for any $x \in G$ it is the case that $x^3 = gxg$.

I've been stumped on this one. All I have found is that $e = e^3 = geg = g^2$. Does anyone have advice on some starting points to solve this?

Thanks

  • It's redundant (and illogical) to write both "for any $x\in G$" and "for all $x\in G$" in the same claim. – anon Feb 19 '18 at 02:05
  • Much more is true. You have $gx^3=xg=gx$, so $x^2=1$, so every element has order 2. – almagest Feb 19 '18 at 15:20

1 Answers1

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Hint: Let $a,b\in G$. Therefore, $(ab)^3=gabg=gag^2bg=a^3b^3$.

TPace
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    From this, I se that $baba=a^2b^2$. What’s the next step? – Lubin Feb 19 '18 at 01:56
  • Using @almagest's comment, all the elements have order $2$. So $baba=e\implies ba=a^{-1}b^{-1}=ab$. –  Mar 14 '20 at 16:35