Why do abstract algebra texts generally define a group something like more-or-less this...
Let * denote a binary operation on a set G.
For all x, y, z in G x*(y*z)=(x*y)*z
There exists an element 1 in G, such that for all x in G, x*1=x
For all x in G, there exists an x' in G, such that x*x'=1.
Instead of say using a definition like this:
Let * denote a binary operation, ' a unary operation, 1 a nullary operation on G (as best as I understand it, a nullary operation can get thought of a mapping from the empty set to G here, since it does hold that "if for all x belonging to the empty set, then there exists exactly one output in G", since the hypothesis in quotes comes as false, this statement always obtains.)
For all x, y , z in G x*(y*z)=(x*y)*z
For all x in G, x*1=x
For all x in G, x*x'=1
I know that S. C. Kleene's Mathematical Logic has a short section where a small few theorems of group theory get proved using something like the second definition of a group, but only a handful, and little else of group theory or any other algebra gets developed. Does anyone know of texts that study some algebraic systems like groups, rings, monoids etc. using definitions more like the second than the first? That is, study algebras not like how universal algebra texts just mention (maybe I haven't read this thoroughly enough though) several algebras and move on, but rather actually study particular algebras using definitions like the second? Additionally, I see that the Schaum's Outline of Group Theory happily uses a fair amount of reverse Polish notation. Does anyone know of any algebra texts that use either a fair amount of Polish notation, or a fair amount of reverse Polish notation?
Thanks for any help here.
Answer
In fact, if you want to define groups as a variety of $\Omega$-algebras, one does in fact define a group this way: as an algebra with signature $(2,1,0)$ and so on.
This lets you fit group theory (and later, ring theory) into the wider tapestry of Universal (or General) Algebra; see for example George Bergman's An Invititation to General Algebra and Universal Constructions.
However, as it happens, there are a lot of things which are true for groups that are not generally true for universal algebras. To consider one example, any semigroup homomorphism between two groups must in fact be a group homomorphism. This is not the case even for monoids (you can have a semigroup homomorphism between two monoids that is not a monoid homomorphism, because it does not map the identity to the identity). If you don't know groups "well enough" (and view them merely as universal algebras with signature $(2,1,0)$ and satisfying appropriate identities), then in order to check that a map between groups is a homomorphism you would need to check that it maps products to products, inverses to inverses, and the identity to the identity. In order to "get away" with just checking that it is a semigroup homomorphism, you would need to prove that this is the case... and this turns out to be essentially equivalent to performing the verification that the definition of group you quote first uniquely determines the identity and the inverses.
So, if they are equivalent, why use one and not the other? A couple of reasons: one is certainly historical inertia. Groups were originally defined only in terms of their binary operation. The second is parsimony: when you have a concept, and it's ubiquituous (as the group concept is) you want your definition to be as parsimonious as possible, because you want to make it easy to verify that a given instance is in fact a group. We could add all sorts of clauses to the definition of groups (clauses which are theorems of the standard definition) that would make further theorems much easier to prove; but that would mean that if you find a set lying on the street and you want to check if it is a group, you would need to check all the extra clauses.
Using the "usual" definition, you only need one operation and three properties. Using the universal algebra definition, you would need to check three operations, and three properties. So you end up having to do more checking to see if what you have before you is indeed a group or not. Generally, it's better to need to do less checking rather than more checking to see if the theory applies.
I don't really know any text that use prefix or suffix notation for the binary operation of a semigroup, so I can't help with the final question.
Since this came up in the comments, let's clarify a few things about "nullary operation."
First, whether you define a function as a set of ordered pairs, or you define it in terms of a domain, a codomain (and regardless of how you define codomain), and a rule that associates to every element of the domain a single element of the codomain, I think we can agree that if $f$ and $g$ are functions with the same domain and same codomain, then $f=g$ if and only if for every element $x$ in the domain, $f(x)=g(x)$.
So, if $A$ is a set, how many distinct functions are there from $\emptyset$ (the emptyset) to $A$? As you note, the condition of being a function (for every element of the domain there is one and only one element of $A$) is satisfied vacuously, so it seems like you can just take your function to be "whatever", and it will satisfy the definition.
But the point I raised in comments was that in point of fact, there is only one function because of what equality of functions means. If $f$ and $g$ are two functions with domain $\emptyset$ and codomain $A$, then $f=g$ by vacuity: for every $x\in\emptyset$, we have $f(x)=g(x)$. (Put another way, for $f$ and $g$ to be different, there would have to be an element in the empty set where $f$ and $g$ disagree; no such element, so they aren't different). That means that any two functions with domain $\emptyset$ and codomain $A$ are necessarily equal as functions. So there is one and only one function with domain $\emptyset$ and codomain $A$.
If you define a function from $X$ to $Y$ as a subset of $X\times Y$ (ordered pairs) that satisfies certain properties, then it turns out that $\emptyset$, as a function from $\emptyset$ to $A$ (as a subset of $\emptyset\times A = \emptyset$) satisfies this definition, so the function from $\emptyset$ to $A$ is $\emptyset$, the "empty function." That's why I said that there is one and only one function from the empty set to $A$.
Now, about $n$-ary (and $0$-ary or nullary) operations: generally, an $n$-ary operation on $A$ is defined to be a function from $A^n$ to $A$. So a nullary operation is a function with domain $A^0$ and codomain $A$.
What is $A^0$? Well, here it is useful to define arbitrary cartesian products: if $\{A_i\}_{i\in I}$ is a family of sets, then $\mathop{\times}\limits_{i\in I}A_i$ is defined to be the set of all functions $f\colon I\to \cup A_i$ such that $f(i)\in A_i$. For $I=\{1,2,\ldots,n\}$, this can be naturally seen to be "the same" as the idea of $n$-tuples: the $n$-tuple $(a_1,\ldots,a_n)$ is identified with the function that maps $i$ to $a_i$; and a function $f\colon I\to \cup A_i$ with $f(i)\in A_i$ can be associated with the tuple $(a_1,\ldots,a_n)$.
So instead of defining $A^n$ as the set of $n$-tuples, it makes more sense to define it as the product $\mathop{\times}\limits_{i=1}^n A$; this allows for easy generalization to other sets: for any set $I$, you can define an $I$-tuple of elements of $A$ simply by $\mathop{\times}\limits_{i\in I} A$. This is
$$\mathop{\times}_{i\in I} A = \bigl\{ f\colon I\to A\bigm| f\text{ is a function}\bigr\}$$
the set of all functions from $I$ to $A$. By analogy to $A^n$, we write this as $A^I$.
Using this notation, $A^0$ is the set of all functions from $0$ to $A$; under the usual definition of the natural numbers as sets, we have $0=\emptyset$, $1=\{0\}$, $2=\{0,1\}$, etc. So $A^0$ is
$$A^0 = \{ f\colon 0 \to A\mid f\text{ is a function}\} = \{f\colon\emptyset\to A\mid f\text{ is a function}\}.$$
But we just talked about this. There is one and only one function with domain $\emptyset$ and codomain $A$; so
$$A^0 = \{f\colon \emptyset\to a\mid f\text{ is a function}\} = \{\emptyset\}.$$
So, a nullary operation on $A$ is a function $A^0\to A$. But $A^0 = \{\emptyset\}$. So a nullary operation on $A$ is a function $\{\emptyset\}\to A$. There is a natural correspondence between functions from a singleton to $X$ and the elements of $X$: a function $f\colon\{a\}\to X$ corresponds to the element $f(a)$; and an element $b\in X$ corresponds to the function that sends $a$ to $b$. So the nullary operations on $A$ are in one-to-one correspondence with the elements of $A$, which is why nullary operations are sometimes called "distinguished elements of $A$": you can think of a nullary function as its unique value.
In any case, a nullary operation is not a mapping form the empty set to $G$, but rather it's a mapping from the set of all mappings from the empty set to $G$, to $G$: it's a function $G^{\emptyset}\to G$.
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