Query on algebraic systems, varieties and free objects

Question

For many purposes one usually consider algebraic systems that are groups with respect one of their operations and $f^n(1,\dotsc, 1) = 1$ for all operations where $1$ is the neutral element with respect to chosen group operation (This is quoted from the book "$p$-automorphisms of finite $p$-groups by E.I.Khukhro).

I was wondering how there could be only one operation even in the perspective of systems contain "group" objects. The variety of all groups require three operations (i) product being a binary operation, (ii) inversion being unary and (iii) assigning identity $1$ being a $0$-ary operation and there are laws connecting these operations. This need to be bare minimum. How can one unify all three operations?

I would appreciate if anyone point out if I misunderstood the meaning of the first paragraph above.

Answer 1

I think a pretty common misconception with beginning universal algebraists is that the definition of groups with 3 fundamental operations is mandatory, rather than a personal preference. You might not belong to that category of people, but your question seems similar.

You only need one operation in the definition of a group:

A group is a pair $(G,\cdot)$ where $G$ is a set and $\cdot$ is a binary operation on $G$ that satisfies the following:

1. $\forall a,b,c\in G,[ a\cdot(b\cdot c)=(a\cdot b) \cdot c]$
2. $\exists a \in G, [(\forall b\in G, a\cdot b = b)$ and $(\forall b, \exists c\in G, c\cdot b =a)]$

This fully defines groups. It follows from these axioms that the element whose existence is provided in axiom 2 is, in fact, unique. So, to make it the definition easier to read, we can add that unique element to the signature:

A group is a triple $(G,\cdot,1)$ such that $G$ is a set, $\cdot$ is binary operation on $G$, and $1$ is an element of $G$ that satisfies the following:

1. $\forall a,b,c\in G,[ a\cdot(b\cdot c)=(a\cdot b) \cdot c]$
2. $\forall b\in G, [1\cdot b=b]$
3. $\forall b\in G, \exists c\in G, [c\cdot b =1]$

The problem (for universal algebraists, at least) with this axiom system is that $(\mathbb{N},+, 0)$ is a subalgebra (meaning it is closed under $+$ and contains $0$) of the group $(\mathbb{Q},+,0)$, but $(\mathbb{N},+,0)$ is not a group. We want subalgebras of groups to be groups.

The reason why general definitions of subalgebra, homomorphism, and product do not behave well with this axiom system is because the axiom system does not consist solely of universally quantified equations.

By adding the unary operation $x\mapsto x^{-1}$ to the signature, we can make sure that each of the axioms is a universally quantified equation and that the class of all groups is a variety (i.e. closed under subalgebra, homomorphic image, and arbitrary direct products). With the other axiom systems, the class of groups was just a first-order-defined subclass of the class of all semigroups (or monoids).

Answer 2

The only problem I see is that existence of an element $a$ satisfying axiom $2$ forces to keep a $0$-ary operation a priory. The suggestion by Max probably bypass this without keeping the operation. So here is my suggestion : A group $G$ is an algebraic system with a binary operation ${\cdot}/{\cdot} : G \times G \rightarrow G$ that satisfy the laws :

(i) $(a/((a/a)/b))/((a/a)/c) = a/((a/a)/(b/((a/a)/c))))$ for every $a,b,c \in G$,

(ii) $(a/a)/((a/a)/b) = b = b/(a/a)$ for every $a,b \in G$,

(iii) $(a/a) = (b/b)$ for every $a, b \in G$.

Now the existence of identity and inverse would follow from this.