To show that the endomorphism ring of any locally cyclic group is commutative

Question

Let $G$ be a locally cyclic group , then this wiki page https://en.wikipedia.org/wiki/Locally_cyclic_group claims that the endomorphism ring $End(G)$ is commutative , but I am unable to see how it is ... this would also solve this problem If $G$ is a locally cyclic group , then is $\operatorname{Aut}(G)$ abelian? ( but independent answers there are also appreciated ) . Please help . Thanks in advance

Answer 1

Here's an elementary proof.

Let $\alpha$ and $\beta$ be endomorphisms of $G$ and $g\in G$. So we need to show $\alpha\beta(g)=\beta\alpha(g)$.

Since $G$ is locally cyclic, the subgroup generated by $g,\alpha(g),\beta(g)$ is cyclic, generated by $s$, say.

In turn, the subgroup generated by $s,\alpha(s),\beta(s)$ is cyclic, generated by $t$, say.

Replacing $s$ by the smallest positive power of $t$ that is in $\langle s\rangle$, which is another generator of $\langle s\rangle$, we can assume that $s=t^l$, and that $t^i\in\langle s\rangle$ if and only if $l$ divides $i$.

Let $g=s^k$, $\alpha(s)=t^m$ and $\beta(s)=t^n$.

Then $\beta(g)=\beta(s)^k=t^{nk}$. Since $\beta(g)\in\langle s\rangle$, $nk$ is divisible by $l$ and $\beta(g)=s^{nk/l}$. So $\alpha\beta(g)=\alpha(s^{nk/l})=t^{mnk/l}$.

But similarly $\beta\alpha(g)=t^{mnk/l}$, so $\alpha\beta(g)=\beta\alpha(g)$.