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I was trying to suppose for a contradiction that it is abelian, then: $$ba^3 = a^2b = ba^2\implies a = 1$$ Then we have $G = \langle a,b\mid a = 1\rangle$, which I believe is isomorphic to $\mathbb{Z}$, but I cannot get a contradiction from here. Could someone give me some hints on where to start this problem, I would really appreciate that!

Thank you :)

Shaun
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Paimonium
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    It is enough to find a group with two elements $a$ and $b$ that do not commute, and which satisfy the relation. The relation can be rewritten as $b^{-1}a^2b = a^3$. So find a group with an element whose square is conjugate to its cube. – Arturo Magidin Aug 08 '23 at 03:58
  • $a = 1$ is the contradiction you seek. $G$ is presented as a group generated by two elements, $a$ and $b,$ such that they satisfy the relation $a^2b = ba^3.$ By definition, none of the generators in a presentation of a group can be the identity element of the group. – Edd Aug 08 '23 at 04:01
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    Consider the dihedral group $$D_5=\left<A=\begin{pmatrix}\cos(2\pi/5)&-\sin(2\pi/5)\\sin(2\pi/5)&\cos(2\pi/5)\end{pmatrix},B=\begin{pmatrix}1&0\0&-1\end{pmatrix}\right>.$$ It can be verified that $BA^3=A^2B$, hence there exists a surjective homomorphism $\phi:G\to D_5$. However, since $D_5$ is nonabelian, the same must be true for $G$. – Sangchul Lee Aug 08 '23 at 04:19
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    Because a definition of a group by generators and relations is the universal group satisfying those relations, it suffices, as Arturo Magidin explained, to find an example of non-commuting elements $a,b$ of any group that satisfy the given relation. One way to achieve that (similar to Sangchul Lee's suggestion) is to pick a $p$-cycle $a$ from a symmetric group $S_n, n\ge p$, $p>3$ a prime. Then $a^2$ and $a^3$ are also $p$-cycles, and hence conjugate by some element $p$. If $a$ and $b$ were to commute, so would $b$ and $a^2$. But $a^2$ and $a^3$ distinct permutations so they cannot. – Jyrki Lahtonen Aug 08 '23 at 04:34
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    @Edd You are wrong when you say that none of the generators can be the identity. That is not part of the definition of a presentation. – ancient mathematician Aug 08 '23 at 06:25
  • ^ ...hence conjugate by some element $b\in S_n$... A typo in my earlier comment. – Jyrki Lahtonen Aug 08 '23 at 08:00
  • @Edd here is an example of a presentation of the trivial group where one of the relations is the relation appearing in the question above: https://math.stackexchange.com/questions/66573 – KCd Aug 08 '23 at 10:04
  • @Edd and here is another nonobvious presentation of the trivial group: https://math.stackexchange.com/questions/1023341 – KCd Aug 08 '23 at 10:09
  • @KCd I stand corrected. Sorry, I never actually encountered a problem in which the set of relations is "too big" for the generators, so that the quotient of the free group "loose" some of them, or even worse, it becomes trivial. Thanks for the correction. – Edd Aug 08 '23 at 11:50

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The Baumslag-Solitar group $B(m,n)$ is the group given by the presentation $$ BS(m, n) = \langle a, b \mid b^{-1}a^m b = a^n \rangle $$ where $m,n \in \mathbb{Z} \setminus \{0\}$.

For your group $G$ we have $G=B(2,3)$ by definition, i.e., $G$ is the Baumslag-Solitar group with $m=2$ and $n=3$. This group is not even virtually abelian. Actually, $B(m,n)$ is abelian if and only if $mn=1$. A proof has been given, among other posts, here:

Condition for Baumslag-Solitar group to be virtually abelian

Dietrich Burde
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