For $x,y$ in G, prove that $x = y =e$ if $xy^2=y^3x $ and $yx^2=x^3y$ where $e$ is the identity element in G.

I have filled pages trying to solve but unable to reach the solution. Also, Is there any general method to approach such questions?

  • $\begingroup$ There is no general method. The general problem is called the word problem (for groups), and it is considered hard. $\endgroup$ – Arthur Jan 3 '18 at 6:34
  • $\begingroup$ Does that mean I have to mug up approach for these questions? Most of the time I end up with the given equations when trying to solve. $\endgroup$ – LoneCuriousWolf Jan 3 '18 at 6:39
  • $\begingroup$ Is $G$ a finite group? $\endgroup$ – N. S. Jan 3 '18 at 7:02
  • $\begingroup$ It's not given as such. $\endgroup$ – LoneCuriousWolf Jan 3 '18 at 7:03
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    $\begingroup$ An essential duplicate. Mind you, I like 1Enigma1's solution over mine (Steve D's is better than mine). Using the fact that conjugates have the same order saves a number of steps. Therefore I won't cast the first vote to close as a dupe. $\endgroup$ – Jyrki Lahtonen Jan 3 '18 at 7:30

Let $xy^2=y^3x$ say $(1)$ and $yx^2=x^3y$ say $(2)$

From the relation $(1)$, we have $xy^2x^{−1}=y^3$ say $(3)$.

Computing the power of $n$ of this equality yields that $xy^{2n}x^{−1}=y^{3n}$ for any $n\in \mathbb{N}$.

In particular, we have $xy^4x^{−1}=y^6$ and $xy^6x^{−1}=y^9$.. Substituting the former into the latter, we obtain $x^2y^4x^{−2}=y^9,$ say $(4)$

Cubing both sides gives $x^2y^{12}x^{−2}=y^{27}$.

Using the relation $(3)$ with $n=4$, we have $xy^8x^{−1}=y^{12}$ Substituting this into equality $(4)$ yields $x^3y^8x^{−1}=y^{27}$

Now we have $$y^{27}=x^3y^8x^{−1}=(x^3y)y^8(y^{−1}x^{−3})=yx^2y^8x^{−2}y.$$

Squaring the relation $(4)$, we have $x^2y^8x^{−2}=y^{18}$. Substituting this into the previous, we obtain $y^{27}=y^{18}$, and hence $$y^9=e$$ Note that as we have $xy^2x^{−1}=y^3$, the elements $y^2,y^3$ are conjugate to each other. Thus, the orders must be the same. This observation together with $y^9=e$ imply $y=e$.

It follows from the relation $(2)$ that $x=e$ as well.

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    $\begingroup$ Can I ask what was the background idea(s) you had to find this!?! $\endgroup$ – mathcounterexamples.net Jan 3 '18 at 9:06

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