Centre of the group with presentation $\langle a, b | a^2 = b^3 \rangle$ I'm trying to find the centre of the group with presentation $\langle a, b | a^2 = b^3 \rangle$.  Not really sure how to get started - any help appreciated!
 A: A general approach is the following. I've included an approach that is more specific to this example at the end.
Firstly, realise that your group has a relative presentation of the form $G=\langle H, K; a=\phi(a)\rangle$ where $\phi: A\mapsto B$ is an isomorphism between subgroups $A$ and $B$ of, respectively, $H$ and $K$. Such a construction is called a free product with amalgamation, and the groups $H$ and $K$ are called the factor groups. We write $G=H\ast_{A=B}K=H\ast_CK$, where $C$ is the common subgroup (in our example, $C=\langle a^2\rangle=\langle b^3\rangle$). Then open the book Combinatorial Group Theory by Magnus, Karrass and Solitar. This is the first handbook of combinatorial group theory, the second being the book by Lyndon and Schupp of the same name (but Lyndon and Schupp do not have the theorems relevant to what I am about to say).
Now, Theorem $4.5$ of MKS states the following.
Theorem: Let $G=H\ast_CK$ be a free product with amalgamation. Suppose $g, h\in G$ such that $gh=hg$. Then one of the following happens.


*

*$g$ is in a conjugate of $C$.

*$h$ is in a conjugate of $C$.

*Neither $g$ nor $h$ is in a conjugate of $C$ but $g$ is in a conjugate of a factor. In this case, $h$ is in that same conjugate of a factor.

*Neither $g$ nor $h$ is in a conjugate of $C$ but $h$ is in a conjugate of a factor. In this case, $g$ is in that same conjugate of a factor.

*Neither $g$ nor $h$ is in a conjugate of a factor. In this case, $g=U^{-1}cU\cdot W^i$ and $h=U^{-1}c^{\prime}U\cdot W^j$ where $U, W\in G$, $c, c^{\prime}\in C$ and $U^{-1}cU, U^{-1}c^{\prime}, W$ pairwise commute.


So, suppose $g\in Z(G)$ and consider the element $h:=a$. As $a$ is in a factor, one of the first four cases must happen, so $g=W^{-1}a^iW$ for some word $W$. Taking $h:=b$ and applying the same logic we see that $g=U^{-1}b^jU$. As factor groups are not conjugate, we see that $a^i, b^j\in C$ (so $i=2i_0$ and $j=3j_0$). However, we know (see NickyHester's comment) that $C$ is contained in the centre of $Z(G)$, $C\leq Z(G)$, and so $g=a^i=b^j$. Therefore, $g\in C$. Thus, $C=Z(G)$.
Now, for this specific example this theorem is a bit heavy-handed. If I were you I would try to prove the result using the "normal form" (it is not a normal form, but anyway...) for free products with amalgamation (you can find this in MKS or Lyndon and Schupp). So, every element which is not contained in a factor group has the following form where $i_1$ and $j_m$ are possible zero, $i_p=1$ and $j_p\in \{1, 2\}$.
$$
a^{i_1}b^{j_1}\ldots a^{i_m}b^{j_m}
$$
Try to slide an $a$ and a $b$ past such a word. What happens?
