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Consider a copy of $\mathbb{Z}$

This forms a group when equipped with the "+" operator

Let us consider now $\text{End}(\mathbb{Z},+) $ the set of endomorphisms that preserve the additive structure, these endomorphisms are functions and have an associative structure that they "fit" into, namely

$$ (\text{End}(\mathbb{Z}, +), \times ) $$

Forms a set with an associative operator.

We can go further and look at the "closure" of

$$ (\text{End}(\mathbb{Z}, +) - \lbrace (x \rightarrow 0*x) \rbrace, \times ) $$

By defining inverses for each of the endomorphisms and this set is equivalent to $\mathbb{Q} - \lbrace 0 \rbrace $.

What is really weird to me, is if I repeat this algorithm, of taking a group, and looking at the endomorphisms of the group, and "completing" the endomorphism structure (and removing some elements) I don't end up with another group.

Exponentiation is neither associative nor commutative.

Why abstractly is this happening? For example if I have some group G and I look at

$$ G \rightarrow \text{End}(G) \rightarrow \text{End}^2 (G) \rightarrow ... $$

And G is commutative, and End(G) also forms a natural commutative group, then why does $\text{End}^2(G)$ suddenly not form a commutative and associative group, or to relax our constraints maximally, when and why do these endomorphism sequences stop creating semigroups?

Some notes: (in response to @Qiaochu’s answer)


I think our viewpoints are subtly different. If we take the endomorphisms of an object S we can identify each endomorphism with an element of S (often by seeing how the endomorphism moves a non identity element). Having built a map $m: S \rightarrow \text{End}(S)$ we can then define the binary operator $B: S \times S \rightarrow S$ by $B(a,b) = m(a)[b]$. It is the case that when S was the integers with Addition then B was commutative and associative (since B was multiplication). But when S is the rationals (without 0) with multiplication then B is exponentiation and exponentiation is neither commutative nor associative. Basically I’m trying to understand what properties does a group S need to have so that the B operation constructed as above is at the least associative

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    $\begingroup$ The question is where your map $m:S\to \operatorname{End}(S)$ comes from. Usually there isn't such a map, and I think the process by which you are going from addition to multiplication to exponentiation is a lot less systematic than you suggest. $\endgroup$ – Eric Wofsey Mar 8 at 23:47
  • $\begingroup$ I agree with Eric. This is not the right way to think about exponentiation. $\endgroup$ – Qiaochu Yuan Mar 9 at 3:02
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What is really weird to me, is if I repeat this algorithm, of taking a group, and looking at the endomorphisms of the group, and "completing" the endomorphism structure (and removing some elements) I don't end up with another group.

I don't know what construction you're describing here. The endomorphisms of $\mathbb{Q}^{\times}$, like the endomorphisms of any abelian group, form a ring (more or less a ring of infinite matrices). Unlike $\mathbb{Z}$ this ring is noncommutative and has zero divisors, so it's not at all clear what it would mean to take fractions.

In general,

  • the endomorphisms of any object in any category whatsoever always form a monoid under composition,
  • for abelian groups we have the special property that $\text{Ab}$ is canonically enriched over itself, so endomorphisms form a monoid internal to $\text{Ab}$, which is to say a ring. Generally this ring will be noncommutative and have zero divisors.
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  • $\begingroup$ My comment is too long, I’ve made an addendum to the Question, please review and let me know your thoughts :) $\endgroup$ – frogeyedpeas Mar 8 at 23:39

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