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So I hear that the endomorphism ring of an abelian group is not always commutative. In particular, I'm looking at the abelian group $A=\mathbb{Z}\times\mathbb{Z}$, and considering $\text{End } A$. I can't find a counterexample to show that $\text{End } A$ is not commutative. Does anyone know of two such endomorphisms that don't commute?

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Can you find two matrices with integer entries for which $AB \neq BA$? – t.b. May 21 '11 at 6:59
If you pick two endomorphisms of $\mathbb Z$ you will find they do commute (do do this). So let's do the “next” case: if you pick a pair of endomorphisms of $\mathbb Z\oplus\mathbb Z$ more or less however you want, you will find that they don't commute. It is just a matter of trying until you find an example. You should try! – Mariano Suárez-Alvarez May 21 '11 at 19:46

Hint: try to show that the ring $M_2(\mathbb{Z})$ of $2 \times 2$ matrices with $\mathbb{Z}$-coefficients acts effectively faithfully on $\mathbb{Z}^2$ by endomorphisms. This is a non-commutative ring, so this will answer your question.

More generally, if $A$ is an abelian group and $n \in \mathbb{Z}^+$, then one can show that $\operatorname{End}(A^n) = M_n(\operatorname{End}(A))$. (It happens that the endomorphism ring of the group $\mathbb{Z}$ is canonically isomorphic to the ring $\mathbb{Z}$, so the previous paragraph is a special case of this.) This gives many examples of non-commutative endomorphism rings.

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Thanks Pete, what do you mean "acts effectively"? Following @Theo Buehler's comment, I found A=[0 1// 2 0] and B=[0 2 // 1 0] to be $2\times 2$ matrices which do not commute, but how does this help exactly? – Dom May 21 '11 at 7:09
I mean that there is an injective homomorphism $M_2(\mathbb{Z}) \rightarrow \operatorname{End}(\mathbb{Z}^2)$. In other words, distinct matrices give rise to distinct endomorphisms. – Pete L. Clark May 21 '11 at 7:29
@Dom: you may have seen this called acting "faithfully", rather than "effectively". – Chris Eagle May 21 '11 at 9:57
Thanks, @Chris. Both terms are widely used but upon reflection I prefer faithfully for ring actions. – Pete L. Clark May 21 '11 at 20:03

Every ring acts by left multiplication as endomorphisms of its underlying abelian group ("Cayley's theorem for rings"), so if you believe that there exist non-commutative rings, then you already believe that there exist non-commutative endomorphism rings.

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I love the way you phrased that. – Jonas Meyer May 21 '11 at 7:33

I know other people have given good answers but here's an explicit example: $\phi : (a,b) \mapsto (b,a)$ and $\theta: (a,b) \mapsto (a+a,b)$ both are homomorphisms from $A$ to $A$, but the order you apply them matters.

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An endomorphism of $\mathbb{Z}\times\mathbb{Z}$ can be represented as a $2\times2$ matrix with entries in $\mathbb{Z}$, and all such matrices give endomorphisms of $\mathbb{Z}\times\mathbb{Z}$. (The ones with determinant 1 give automorphisms.)

So any two $2\times 2$ matrices that don't commute will provide you with an example.

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Thanks, I think this answered my question in Pete L. Clark's answer above. – Dom May 21 '11 at 7:11

HINT $\rm\ \ \ a\ x + y\:\ \ne\:\ a\ (x+y)\ \ $ if $\rm\ \ a\ne 1$

$\quad\ $ i.e. $\rm\ \ \ \ \ (x,\:y)\ \mapsto\ (a\:x,\:y) $

$\quad$ and $\rm\ \ \ \ \:(x,\:y)\ \mapsto\ (x+y,\:y)\ \ $ do not commute.

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