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This seems like a very basic question but got me confused. When defining a group we introduce the unit element $e$ which has the following property $$ge = eg = g \quad \forall g\in G$$

and then the inverse for which we need the unit:

$$gg^{-1} =g^{-1}g = e$$

Is it possible to do it the other way around? Can we construct the unit element from the inverse? I think not but haven't found a convincing argument except that we use the unit in the definition of the inverse. But perhaps it is thinkable to conceive a completely different way.

Shaun
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Jakob Elias
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    If you have not defined $e$ yet, what would be your definition of the inverse element $g^{-1}$? – Edu Apr 28 '17 at 14:02
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    How about $hg^{-1}g=hgg^{-1}=h$, from which you can easily define $e=gg^{-1}$. Uniqueness follows as usual. – MM8 Apr 28 '17 at 14:03
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    In e.g. monoids inverses do not have to exist for every element, but there is an identity. I never met the opposite situation. – drhab Apr 28 '17 at 14:04
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    I suppose you could do it. A common similar reduction is to note that you only need one operator, $g\star h = g\cdot h^{-1}$. Then you can state your axioms in terms of $e=h\star h$, $h^{-1}=e\star h$ and $gh=g\star h^{-1}$. But the axioms become quite noisy, and there are deep reasons we like to talk about associative operations. – Thomas Andrews Apr 28 '17 at 14:10
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    Not quite an answer, but close: One can define groups by axioms which characterise a binary operation $a/b$ which turns out to be (in the standard notation) $a/b:=ab^{-1}$. The identity and products will emerge nicely. – ancient mathematician Apr 28 '17 at 14:13
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    @ThomasAndrews beat me to it. – ancient mathematician Apr 28 '17 at 14:14
  • The axioms for the $a/b$ operation (to be interpreted as $a b^{-1}$) can be massaged to be less noisy than a straightforward translation of the group axioms: $a/a = b/b$, $a/(a/a) = a$, $(a/c)/(b/c) = a/b$ (and then you also need to assume the set is nonempty). – Daniel Schepler Apr 28 '17 at 23:44
  • As an aside, the traditional definition of group is more about answering the question "I have an operation; is it a group operation?" than about answering the question "What is a group?" When answering the latter question in a similar fashion, you would posit three operations ($x \cdot y$, $x^{-1}$, and $e$) and replace all of those existential claims with identities asserting that $e$ is an identity and that $x^{-1}$ is the inverse of $x$. –  Apr 29 '17 at 00:11
  • It's actually kind of weird that both approaches give the same theory of groups. Monoids, for example, demonstrate the divergence; we require that monoid homomorphisms preserve the identity, which means it is part of the structure of being a monoid, as opposed to monoids simply being semigroups with the property of having an identity. –  Apr 29 '17 at 00:14

4 Answers4

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I suppose you could do it.

A common similar reduction is to note that you only need one operator, $g\star h = g\cdot h^{-1}$. Then you can state your axioms in terms of $e=h\star h$, $h^{-1}=e\star h$ and $gh=g\star h^{-1}$. But the axioms become quite noisy, and there are deep reasons we like to talk about associative operations.

So if you have a set $G$ with a binary operation $\star$ with the following properties:

  1. $G$ is not-empty
  2. For all $h,g\in G$, $h\star h=g\star g$. From here on, we'll write $e=h\star h$.
  3. $e\star(e\star h)=h$, $h\star e=h$.
  4. $h_1\star (h_2\star h_3)=(h_1\star (e\star h_3))\star h_2$.

Once you have such an operation, you can define d $g^{-1}=(g\star g)\star g$ and $g\cdot h=g\star h^{-1}=g\star((h\star h)\star h)$.

One tricky thing is that, without the requirement for an identity, you are going to need to assert that $G$ is non-empty.

There is a deep theoretical reason that we prefer to talk about associative operations first, however. The most fundamental associative operation is function composition. Let $X$ be a set, and let $([X\to X],\circ)$ be the set of all functions from $X$ to itself, with the operation $\circ$ being function composition. $\circ$ is an associative operation.

Turns out, if $(S,\times)$ is any set with an associative operation, then it is equivalent (isomorphic) to some sub-algebra of an $([X\to X],\circ)$ for some set $X$. (You can always use $X=S\sqcup\{I\}$, in fact.) Such "representations" of $(S,\times)$ are a deep common fact in a lot of mathematics, which is related to something called "category theory."

This also indicates why the identity is more primal than inverses.

$([X,X],\circ)$ always has an identity (though $(S,\times)$ might not). So it is easy to "add" an identity to $S$.

However, if $|X|>1$, some elements of $[X,X]$ have no inverses, and if you try to add inverses and keep the associative rule, you end up doing something way more complicated than merely "adding" elements to $S$.

Thomas Andrews
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  • I don't understand, does this definition of $(G,\star)$ induce a group structure $(G,\cdot)$, How is $x\cdot y$ defined by $\star$? How is $x^{-1}$ defined? – miracle173 Apr 28 '17 at 15:16
  • @miracle173 $e$ is defined as $h\star h$, for some $h$. By the first axiom, that is the same no matter which $h$ is chosen. Then you can define $g\cdot h = g\star(e\star h)$ and $g^{-1}=e\star g$. I've added this to the answer. – Thomas Andrews Apr 28 '17 at 15:22
  • Homework: If $(G,\star)$ satifies 1 to 4 and $e$ and $^{-1}$ are defined by $\star$ as described above, proof that $(G,\cdot, e, ^{-1})$ is a group. – miracle173 Apr 28 '17 at 15:34
  • I'm not even 100% sure that is true - I might have missed an axiom. But the above is the idea. The key is that it is sort of unpleasant. In particular, (4) is deeply ugly compared to the associative rule. – Thomas Andrews Apr 28 '17 at 15:38
  • I am also not sure if it is already a group. But what seems more unsatisfactory to me is that I am not sure if you really don't use the neutral element to define the inverse. Actually already in step two you define something that will be the neutral of $\cdot$ operation, even if you would avoid to call it $e$ this uniquely defined element ist the neutral. So I don't understand why you think you construct the unique element from the inverse. – miracle173 Apr 28 '17 at 15:55
  • The induced $\cdot$ is actually not a group. This can be seen by defining $g\star h=e, \forall g,h \in G$. This induces $g\cdot h=e, \forall g,h \in G$, which is not invertible. – miracle173 Apr 29 '17 at 08:26
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Is it possible to do it the other way around?

Sort of, yes. A semigroup $S$ is a group if and only if for all $a\in S$, $$aS=S=Sa.$$ This is sort of like having inverses first. I suppose it's worth noting that this definition uses neither the identity nor inverses.

However, one might say that to define an inverse in the first place, you need an identity.

Having said that, a non-empty, regular semigroup is a group if and only if it is cancellative: Conditions for a regular semigroup to be a group..

Shaun
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The idea of inverse elements, as such, logically requires the notion of an identity element with respect to which inverses can be defined, but there are partial results in the direction you suggest.

One way to go at this is to define, for a semigroup (associative magma) $S$, a quasi-inverse of an element $a\in S$ to be an element $a^\ast\in S$ for which $aa^{\ast}a = a$. It is then the case that if every element of $S$ has an unique quasi-inverse, then $S$ is a group. We choose an element $a$ in $S$ and show that $aa^\ast$ is, in fact, an identity element for $S$ and $x^\ast$ is an inverse for each $x\in S$. (Uniqueness of quasi-inverses is essential here, as the set of all square matrices over a field has non-unique quasi-inverses under matrix multiplication, but is clearly not a group.)

James
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You can replace the axiom of the neutral and the inverse element by the following axiom:

For all $a,b \in G$ each of the following equations has a unique solution. $$ax=b$$ $$xa=b$$ From this axiom the existence of a neutral element and the inverse elements can be deduced.

miracle173
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