showing that a group of order 45 is abelian I'm trying to understand the following proof from Dummit & Foote (pg. 137) which shows why a group of order 45 is abelian. I understand everything but the last two sentences. Why is it that $|G/C_G(P)|=1$? Also, is it always true that $G/Z(G)$ is abelian? Thank you!


*

*They have shown that $\lvert G/C_G(P) \rvert$ is either $1$ or $5$, and then that it divides $48$, so it has to be $1$.

*$C/Z(G)$ need not be abelian in general. But if it is cyclic, as in this case (as it has order a divisor of $5$), then $G = Z(G)$ is abelian. The standard argument goes like this. Suppose $G = Z(G) \langle a \rangle.$ Write two arbitrary elements of $G$ as $z a^{i}$ and $w a^{j}$, with $z, w \in Z(G)$. Then
$$
(z a^{i}) \cdot ( w a^{j}) = z w a^{i} a^{j} = w z a^{j} a^{i} = (w a^{j}) \cdot (z a^{i}).
$$

