What is an example of group homomorphisms $f,g: M \to M$ where $M$ is a simple abelian group such that $f\circ g \ne g\circ f$ ?
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See https://math.stackexchange.com/questions/34449/automorphism-group-of-an-abelian-group – Jakob Hansen Jun 28 '16 at 16:59
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An abelian group is simple iff it is finite of prime order $\;p\;$ , and in that case its automorphism group is cyclic of order $\;p-1\;$ , so all its automorphisms commute with each other.
If $\;f\;$ is an endomorphisms of such a cyclic group of order a prime then it is either the trivial homomorphism or the identity one (as it has no non-trivial subgroups), and again we have commutativity.
DonAntonio
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Considering $\mathbb{Z}_3 = {0,1,2}$, a simple abelian group. Although $End(\mathbb{Z}_3)$ is a commutative ring, a group endomorphism $f : \mathbb{Z}_3 \to \mathbb{Z}_3$ determined by $f(1) = 2$ would be an endomorphism not necessarily and identity right? – MathsMy Jun 28 '16 at 18:14
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@Dosomemaths Nop: $$f(1)=2\implies f(2)=4=1;,;;f(0)=0$$ so $;f;$ in fact is na automorphism. – DonAntonio Jun 28 '16 at 20:54
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Yes, it is an invertible endomorphism (automorphism) from $\mathbb{Z}_3$ to itself but is not an identity map. – MathsMy Jun 29 '16 at 02:55
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Let $M=\mathbb{R}$. There exist (uncountably many) functions $f:\mathbb{R}\to\mathbb{R}$ which satisfies $f(x+y)=f(x)+f(y)$ but not linear (to prove it, we use Hamel basis, basis of $\mathbb{R}$ as $\mathbb{Q}$-vector space). Although this functions are not constructible (use AC), I think $f\circ g\neq g\circ f$ for appropriate Hamel functions $f, g:\mathbb{R}\to \mathbb{R}$.
Seewoo Lee
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