Equivalent group presentation I have been trying to determine whether the following groups are isomorphic for a while now with no significant progress.
$G_1 = \langle x, y \mid xyx = yx^2y\rangle$
$G_2 = \langle x, y \mid xyx = yxy\rangle$
I would bet they are not, yet I cannot think of any characteristic that might set them apart. I know that the order of the groups must be the same and that the order of elements must match, but I cannot see how to use this here. In the comments it was suggested that I compute the abelianisations. However, the abelianisation of $G_1$ is $\langle x, y\mid 1=y\rangle$ and the abelianisation of $G_2$ is $\langle x, y\mid x=y\rangle$, which are both infinite cyclic. So abelianisations do not help either.
To show isomorphism I would have to show that there are elements in the other group that obey the same relation and which generate the entire group. I cannot see any obvious candidates.
Any help would be much appreciated.
 A: It is in general a hard (infact, algorithmically undecidable!) problem to determine if two given finite group presentations define isomorphic groups. On the other hand, there are some obvious things to do before you need to start thinking hard.

*

*Compute the abelianisations of each group. This will sometimes tell you that the groups are non-isomorphic.

*Apply some simple Tietze transformations to one or both of the presentations, and check to see if the resulting presentations match. This will sometimes tell you that the groups are isomorphic.

Computing abelianisations doesn't work here - both groups have infinite cyclic abelinisations. So lets try Tietze transformations. Starting with $G_1$:
\begin{align*}
G_1
&\cong \langle x, y \mid xyx = yxy\rangle\\
&\cong \langle x, y, z \mid xyx = yxy, z=xy^{-1}\rangle&\text{add in new generator $z$ as $xy^{-1}$}\\
&\cong \langle x, y, z \mid zy^2zy = yzy^2, zy=x\rangle&\text{prepare to remove the generator $x$}\\
&\cong \langle y, z \mid zy^2z = yzy\rangle&\text{remove $x$ from presentation, tidy up}\\
&\cong\langle x, y \mid xyx = yx^2y\rangle\\
&\cong G_2
\end{align*}
Therefore, the groups are isomorphic.

The above working is formal and boring and not actually what I did. I actually added $z$ and removed $x$ in a single step as follows:
\begin{align*}
G_1
&\cong \langle x, y \mid xyx = yxy\rangle\\
&\cong \langle x, y \mid (xy)y(xy) = y(xy)y, z=xy^{-1}\rangle&x\mapsto xy\\
&\cong \langle x, y \mid xy^2x = yxy\rangle&\text{tidy up}\\
&\cong\langle x, y \mid xyx = yx^2y\rangle&x\leftrightarrow y\\
&\cong G_2
\end{align*}
