A doubt in Bergman's notes

Question

On pg. 8 of these notes, Bergman says that a group $G$ contains an inverse operation $i:G\to G$, along with $\mu:G\times G\to G$ and a "neutral element" $e$. Hence, a group should be referred to as $(|G|,\mu,i,e)$.

Is there really an inverse operation? I thought an inverse is an element such that $\mu(a,a^{-1})=e$.

Is there really any need to mention a group as $(|G|,\mu,i,e)$, when just $(|G|,\mu)$ will do perfectly?

Answer 1

You are correct that since the neutral element and the inverse are unique in a group, they need not be mentioned (they are properties of the multiplication rather than really distinguished elements). But it can still be practical to single out these two things, as they are after all there, and when generalizing the concept of a group to that of a group-element in a category, they become important.

Answer 2

Ok, so saying that $G$ is a group having the operation $\mu$ implicitly gives us an inverse map and a neutral element, so we can just drop $i$ and $e$. But why stop there? Why not just say that saying "$G$ is a group" implicitly gives us an operation, and therefore writing $\mu$ is superfluous? Of course, this is perfectly fine, especially if you never intend to write $\mu$.

The point is that the usefulness of a notation is dictated by the usage the author intends. In something like a universal algebra book, expressing all four pieces of data in compact notation like this is the best thing to serve their needs. In another book, another author may use a slimmed down version because it suits his/her purposes just fine. You can do that provided that the context you're in does not admit chances for confusion.

If you intend to talk about a semigroup $S$ and/or a monoid $M$ in the same context, then you'd probably want to adopt similar notations like $(|S|,\mu_S)$ and $(|M|,\mu_M,1_M)$, and one can tell, at a glance, all of the important specified data.

Answer 3

Notice that when you said

an inverse is an element such that $μ(a,a^{−1})=e$.

you are using the notation $a^{-1}$ without having defined it. What, exactly, does $a^{-1}$ mean? One of the axioms of a group is that

for each element $a$, there exists an element $b$ such that $\mu(a,b) = \mu(b,a) = e$.

By skolemizing this existence statement, we obtain the equivalent statement:

there is a function $i$ with the property that
for each element $a$, we have $\mu(a, i(a)) = \mu(i(a), a) = e$

and we conventionally denote $i(a)$ as $a^{-1}$. This is the definition of the $a^{-1}$ notation.

In informal treatments, we leave all this implicit, but the idea is still there: whenever you write ‘$a^{-1}$’ you are implicitly assuming the existence of a function that maps each $a$ to its inverse. Until you construct such a function, your use of the notation $a^{-1}$ is unjustified.

Answer 4

Yes, there really is an inverse operation in a group. Remember that you don't just need to have a single inverse $x^{-1}$ for some arbitrary $x$, but an inverse for every $x \in |G|$. Given these inverses, we can define a function $\iota: |G| \to |G|$ as $\iota(x) = x^{-1}$, and this $\iota$ is precisely the operation that, applied to any element of the group, gives its inverse.

As for your second question, if you scroll down to page 9, you'll find the following note by Bergman that directly addresses it:

There is another denition of group that you have probably also seen: In effect, a group is defined to be a pair $(|G|,\cdot)$, such that $G$ is a set, and $\cdot$ is a map $|G|\times|G|\to|G|$ satisfying $$ (\forall x,y,z \in |G|)\ (x \cdot y)\cdot z = x \cdot (y \cdot z), \\ \tag{1.2.2} (\exists e \in |G|)\ ((\forall x \in |G|)\ e \cdot x = x = x \cdot e) \land \\ \hspace{9em} ((\forall x \in |G|)\ (\exists y \in |G|)\ y \cdot x = e = x \cdot y). $$

It is easy to show that given $(|G|,\cdot)$ satisfying (1.2.2), there exist a unique operation $^{-1}$ and a unique element $e$ such that $(|G|,\cdot, \,^{-1}, e)$ satisfies (1.2.1). (Remember the lemmas saying that neutral elements and 2-sided inverses are unique when they exist.) Thus, these two versions of the concept of group provide equivalent information. Our description using 4-tuples may seem "uneconomical" compared with one using pairs, but we will stick with it. We shall eventually see that, more important than the number of terms in the tuple, is the fact that condition (1.2.1) consists of identities, i.e., universally quantified equations, while (1.2.2) does not. But we will at times acknowledge the idea of the second definition; for instance, when we ask (imprecisely) whether some semigroup "is a group".