# Example of an endomorphism on an abelian group that is not left multiplication

It is well-known that all endomorphisms on the abelian group ($\Bbb{Z}$,+) can be seen as a left multiplication by some element in some ring structure on ($\Bbb{Z}$,+); namely left multiplication by any integer in the standard $(\Bbb{Z},+,\times)$ ring.

So far every endomorphism on abelian groups that I have examined has turned out to have this same interesting property, but I'm not very knowledgeable in advanced maths.

Can someone provide an example of an abelian group $G$ with an endomorphism that cannot be seen as a left multiplication by some element in some ring structure on $G$?

There are some abelian groups that admit no (possibly nonunital) ring structure with a left unit, and for such groups, the identity endomorphism cannot be multiplication by any element in any ring structure. The standard example of such a group is $G=\mathbb{Q}/\mathbb{Z}$. If you had a ring structure on $G$ with in which (the equivalence class of) $u=a/b$ was a left unit for some $a,b\in\mathbb{Z}$, then $bu=0$, so $bx=b(ux)=(bu)x=0$ for all $x\in G$. But this is clearly false, because (for instance) $b\cdot 1/2b=1/2\neq 0$.
• A-ha! I was hoping it would be $\mathbb{Q}/\mathbb{Z}$. Thanks for the quick response. – Joseph Johnson Dec 27 '15 at 6:50
• May I know why would $bu=0$? Thanks. – yoyostein Jan 20 '18 at 6:38
• $bu$ is the coset of $b\cdot a/b=a$ and $a\in\mathbb{Z}$. – Eric Wofsey Jan 20 '18 at 16:58