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Is there a deeper (categorical) reason for this?

On the one hand a group homomorphism $\phi:(G,\cdot)\to (H,\star)$ preserves 'results of operations' as well as the identity element and inverse elements, but satisfies only one equation: $$\forall g_1,g_2\in G:\phi(g_2\cdot g_1)=\phi(g_2)\star\phi(g_1)$$ while if $\phi$ was a monoid homomorphism instead, it would only preserve the first two things, but need to satisfy the additional equation $$\phi(e_G)=e_H$$ I know how to prove algebraically that a group homomorphism preserves all the mentioned structure, that's not the question. I do not understand why it preserves more structure than the monoid homomorphism, while at the same time having less 'algebraic conditions'.

EDIT: I think what really gives me trouble is that going 'in the natural order' from semigroups to monoids to groups, one starts with one equation for semigroup-homomorphisms, then adds an additional equation for monoid-homomorphisms, and then for group-homomorphisms one goes back to one equation. This seems strange to me.

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5 Answers 5

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I'm not sure about a "categorical" reason, but generally if you restrict to nice subclass of objects, you have less "bad behavior" and so you're theorems/definitions need less restrictions/hypotheses.

As an extreme, what if we just considered the category of "Trivial Groups"? Then every function is a homomorphism! So we don't even need to specify that maps are operation preserving.

An analogy: In general, functions can be 1-to-1 or not, onto or not. But what if we restrict our attention to sets of size $15$? Then a function from a set of size $15$ to a set of size $15$ is 1-to-1 iff it's onto. Thus in my world of size $15$, I can define bijections to be 1-to-1 functions (I get onto for free). The definition of "bijection" is simplified merely because I've moved into a very restrictive world where the phenomena of 1-to-1 and onto are equivalent.

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    $\begingroup$ I like this point of view. We can also look at endomorphisms of vector spaces. If we consider finite dimensional vector spaces, we could define an isomorphism to be an injective endomorphism. In the category of all vector spaces, we have to require the endomorphism to be bijective. $\endgroup$
    – MooS
    Oct 27, 2015 at 20:58
  • $\begingroup$ Exactly......... :) $\endgroup$
    – Bill Cook
    Oct 27, 2015 at 21:00
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    $\begingroup$ Every monoid morphism is a semigroup morphism. So the identity restriction is moving from a wider class of "morphisms" to a special subclass. From this viewpoint we see that in the context of groups there is no distinction between these "generalized" morphisms and regular morphisms. $\endgroup$
    – Bill Cook
    Oct 27, 2015 at 23:17
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    $\begingroup$ You add an equation because you've added structure that you want preserved. Just like going from an abelian group to a ring. Or a ring without identity to a ring with identity. As you add on structure, you usually want your new homomorphisms to preserve that structure. Another example: Every vector space is an abelian group. But abelian group homomorphisms don't preserve scalar multiplication (in general) so we add a condition to restrict our attention to nice homomorphisms (i.e. linear transformations). The point of all of this is sometime when adding structure you so limit your class... $\endgroup$
    – Bill Cook
    Oct 28, 2015 at 16:56
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    $\begingroup$ ...of examples that you don't need extra equations. Groups add on this inversion map: $x \mapsto x^{-1}$. But because the axioms of a group are so restrictive, we don't need to force our homomorphisms to map inverses to inverses - we get it for free. This is just because groups are so much nicer than our previous class of objects (i.e. monoids). Monoids really aren't that much nicer than semigroups (or that's my opinion anyway). $\endgroup$
    – Bill Cook
    Oct 28, 2015 at 16:58
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Groups, as most commonly presented (as a type of 'universal algebra'), have three operations: the identity, the inverse, and the product, and all three.

By the usual convention, a group homomorphism is required to preserve all three types of structure: that is, the natural definition of group homomorphism requires $\varphi(e) = e$, $\varphi(x^{-1}) = \varphi(x)^{-1}$, and $\varphi(xy) = \varphi(x) \varphi(y)$.

The magic is:

Lemma: Every semigroup homomorphism bewteen groups is a group homomorphism.

For some reason I don't really understand, it is typical to introduce things backwards: to define a group homomorphism as a function satisfying the condition needed to be a semigroup homomorphism, and then prove that it satisfies the conditions to be a group homomorphism.

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  • $\begingroup$ All the math is correct in this answer but your claim about history, namely that the most common common way people have presented groups is as having three operations, seems false to me. I remember being taught that a group is a binary operation such that there exists a unit and for each element there exists at least one other element called its inverse, etc. I'm pretty sure that is the most common way to present them, at least in classes and textbooks. $\endgroup$ Oct 30, 2015 at 0:05
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Preserving the operation is a stronger condition for a group operation than for a monoid operation, since the group operation carries more information (namely inverse elements). So it might not really surprise that you have to make sure the identity element is preserved in a monoid, while it is automatic for groups.

Maybe you can compare this to those induction proofs, where you actually proof a stronger statement, but the induction step becomes easier, since the induction hypothesis carries more information.

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In both cases, from $e_{G} \cdot e_{G} = e_{G}$ you obtain $\phi(e_{G}) \cdot \phi(e_{G}) = \phi(e_{G})$. In the group case you can now multiply by $\phi(e_{G})^{-1}$, a move that will not work in general in a monoid.

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Definition: Given $s \in S$, where $(S, \cdot)$ a semigroup,

$s$ is idempotent $\iff$ $s \cdot s = s$

Here's how I think about it:

  1. Semigroup homomorphisms preserve idempotence: Given a semigroup homomorphism $f: S \to T$, if $s \in S$ is idempotent, then $f(s) = f(s \cdot s) = f(s) \cdot f(s)$, i.e. $f(s)$ is idempotent in $T$.

  2. In a group, the only idempotent element is the identity element.

Combining the two facts, we get for free that semigroup homomorphisms always preserve the identity when applied to groups, but not necessarily when applied to monoids.

Monoids can have idempotent elements which are not the identity, groups can't, semigroup homomorphisms preserve idempotence, so they are only guaranteed to preserve the identity for groups.

I claim this answers your question because idempotence is a categorical concept.

Note: Given a semigroup homomorphism $f: M \to S$, when $M$ is a monoid with identity $e_M$, we always have that $f(e_M)$ is an idempotent element in $S$ but that's it. (In particular, if $S$ is also a monoid, then we have a monoid homomorphism if and only if this idempotent element $f(e_M)$ is the identity of $S$.)

This can also be thought of as a (perhaps unnecessary) elaboration of one of the other answers.

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  • $\begingroup$ I guess maybe I should also note that while the corresponding "monoid identity" axiom is redundant for the axioms of a group homomorphism, it is NOT redundant for a group action of a group $G$ on an arbitrary set $X$, en.wikipedia.org/wiki/Group_action#Definition Because group actions can be thought of as a semigroup homomorphism from the group $G$ to the monoid $End(X)$ of all self-functions $X \to X$, the above actually explains why the identity axiom is not redundant for group actions: math.stackexchange.com/a/2878260/606791 $\endgroup$ Jan 4 at 15:41
  • $\begingroup$ Note that the identity axiom for a group action then implies that the submonoid of $End(X)$ that is the image of the semigroup/monoid homomorphism from $G$ to the monoid $End(X)$ of self-functions of $X$ consists only of invertible elements, i.e. is a group, i.e. is a subgroup of the group of all invertible/bijective functions $X \to X$, $Aut(X)$. $\endgroup$ Jan 4 at 15:45

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