In a method for presenting a subgroup we are given A group $$\ G= \bigl\langle\, x, \ y \mid x^2 yxy^3 , \ y^2 xyx^3\,\bigr\rangle$$ So $$\ G/G' = \bigl\langle\, x, \ y \mid x^2 yxy^3 , \ y^2 xyx^3 , [x, \ y] \,\bigr\rangle.$$

Here we notice that $\ y=x$ in $G/G'$ and hence it is isomorphic to $\ C_7$.

We then proceed to obtain a presentation for $G'$ by only using the Schereier transversal of $\ C_7$.

However in another group, namely $$\ G= \bigl\langle\, x, \ y \mid x^3 , \ y^3 , \ (xy)^3\,\bigr\rangle$$ we say that $G/G'$ is isomorphic to $\ C_3 \times \ C_3$ although by since the quotient is abelian, we find from the last relation that $\; y^3 = x^{-3} $, why don't we do the same?

Why in this second example we did not do as the first one although we have the same result that $\; y^3 = x^{-3}\; $ which means $\;y=x $.


Roots are not unique in groups: if $g^n=h^n$ it does not immediately follow that $g=h$. In your specific example, you have that $x^3=y^3=1$ - but the underlying group is precisely the group $C_3\times C_3$, one of the groups of order $9$. It is not cyclic of order $3$, and $x\neq y$. That is: that $x^3=y^{3}$ does not imply $x=y$. As another example, in $\mathbb{R}\setminus\{0\}$ under multiplication, $1^2=1$ and $(-1)^2=1$ but $1\neq -1$ [and you can mess around with the identity in any group - if $g\neq1$ and $g^n=1$ then $g^n=1^n$ but, by assumption, $g\neq 1$...].

More generally, if $g^n=h^n\neq1$ it does not immediately follow that $g=h$. For example, in the Quaternions we have that $i^2=j^2=k^2=-1$, but of course $i$, $j$ and $k$ are distinct elements. As another example, in $\mathbb{R}\setminus\{0\}$ under multiplication, $2^2=4$ and $(-2)^2=4$ but $2\neq-2$.)

Also, in the first example you do not have "$y^3=x^{−3}$ which means $y=x$". You actually have that $x^3=y^{-4}$ and $y^3=x^{-4}$, which implies $x=y$ as follows: $$ \begin{align*} G/G^{\prime} &=\langle x, y\mid x^2yxy^3, y^2xyx^3, [x, y]\rangle\\ &=\langle x, y\mid x^3y^4, y^3x^4, [x, y]\rangle\\ &=\langle x, y\mid x^{-1}x^4y^4, y^3x^4, [x, y]\rangle\\ &=\langle x, y\mid x^{-1}y^{-3}y^4, y^3x^4, [x, y]\rangle\hfill~~~~\text{here, $y^3x^4$ implies $x^4=y^{-3}$, so sub this in }\\ &=\langle x, y\mid x^{-1}y, y^3x^4, [x, y]\rangle\\ &=\langle x\mid x^7\rangle \end{align*} $$


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