Verifying the group axioms

This survey article deals with the question: given a set, and a binary operation, how do we verify that the binary operation gives the set a group structure? This article views the definition of a group as a checklist of conditions.

Define the set and binary operation clearly
First, identify the set clearly; in other words, have a clear criterion such that any element is either in the set or not in the set. For convenience, we'll call the set $$G$$.

Second, obtain a clear definition for the binary operation. The binary operation is a map:

$$*:G \times G \to G$$

In particular, this means that:


 * $$g * h$$ is well-defined for any elements $$g,h \in G$$
 * The value of $$g * h$$ is again an element in $$G$$

Thus, for instance, the operation which sends real numbers $$x,y$$ to $$x^y$$ is not well-defined when $$x$$ is negative and $$y$$ is not an integer; hence, it does not qualify as a binary operation.

Verify associativity
Associativity requires one to pick three arbitrary elements $$g,h,k \in G$$, and show that:

$$g * (h * k) = (g * h) * k$$

There are various strategies for proving this:


 * If $$G$$ is a finite set, this may reduce to checking it on all possible triples of elements in $$G$$
 * If $$*$$ is described by means of a mathematical expression, we may be able to simplify the expressions on both sides in terms of variables $$g,h,k$$, and show that both sides are equal.
 * If $$G$$ is described as a collection of maps from some set $$S$$ to itself, and the binary operation in $$G$$ is by composition of maps, then associativity is automatic because function composition is associative

Find an identity element
An identity element (also called neutral element)is an element $$e \in G$$ such that, for all $$a \in G$$:

$$a * e = e * a = a$$

Again, we have some strategies:


 * If $$G$$ is a finite set, this may reduce to checking by inspection.
 * If $$*$$ is described by means of a mathematical expression, we may be able to simultaneously solve two generic equations of the form $$a * e = a$$ and $$e * a = a$$. Note that we are trying to solve this as an equation in $$e$$ and identity in $$a$$, i.e., it should be true for all $$a \in G$$. After solving the equation. Note further that if solving just one equation already gives a unique solution for $$e$$, we still need to check that that value of $$e$$ works for the other equation as well.
 * If $$G$$ is described as a collection of maps from some set $$S$$ to itself, and the binary operation in $$G$$ is by composition of maps, the identity element is the identity map

Find an inverse map
Next, we need to demonstrate that for every element $$a \in G$$, there exists $$b \in G$$ such that:

$$a * b = b * a = e$$

Again, we have some strategies:


 * If $$G$$ is a finite set, this may reduce to checking by inspection.
 * If $$*$$ is described by means of a mathematical expression, we may be able to solve a generic equation of the form $$a* b = e$$ for $$b$$ in terms of $$a$$
 * If $$G$$ is described as a collection of maps from some set $$S$$ to itself, and the binary operation in $$G$$ is by composition of maps, the inverse of an element is its inverse as a function

NOTE: Due to the somewhat nontrivial fact that monoid where every element is left-invertible equals group, it suffices to actually find a left inverse for every group element; if every element has a left inverse, we automatically get a two-sided inverse. Equivalently, it suffices to find a right inverse for every group element; if every element has a right inverse, we automatically get a two-sided inverse.

In some special cases
In some special cases, we can by-pass checking various conditions for being a group. We discuss two special cases here:

When the binary operation is commutative
When $$*$$ is commutative, then it suffices to find a left identity element (or right identity element), and it suffices to compute just a left inverse (or just a right inverse).

Actually, even for groups in general, it suffices to find just a left inverse, due to the fact that monoid where every element is left-invertible equals group, so we don't really save anything on inverses, but we still make a genuine saving on the identity element checking.

Subset of a group
Suppose $$G$$ is given to be a subset of a group $$K$$, and the binary operation on $$G$$ is the restriction to $$G$$ of the multiplication in $$K$$. Then:


 * We need to verify that the binary operation induces a well-defined binary operation in $$G$$: the product of two elements in $$G$$ is also in $$G$$.
 * We do not need to check associativity of the binary operation, because it holds in $$K$$
 * Instead of trying to find the identity element of $$G$$, we can simply verify that the identity element in $$K$$, actually lies inside $$G$$
 * Instead of trying to compute the inverse map in $$G$$, we can simply verify that the inverse map in $$K$$, sends $$G$$ to within itself.

Quotient of a group by an equivalence relation
Suppose $$G$$ is obtained as the quotient of a group $$K$$ by an equivalence relation. We want to see whether this equips $$G$$ with the structure of a group. In this case, the only thing we need to check is that the equivalence relation is a congruence. In other words, if $$\sim$$ is the equivalence relation, we need to check that:

$$a \sim b, c \sim d \implies ac \sim bd$$

An abelian group
Here is one example. Consider $$G = \mathbb{R} \setminus \{ -1 \}$$ and define, for $$x,y \in G$$:

$$x * y := x + y + xy$$

We want to show that $$(G,*)$$ is a group. (Note: The group is really coming from a concrete realization of the multiplicative formal group law, but we aren't supposed to use that and instead do the proof from first principles).

Verifying that the binary operation is well-defined
First, we check the closure of $$G$$ under $$*$$. Namely, we need to check that if $$x,y \in G$$ then $$x * y \in G$$. Suppose not. Then, we have:

$$x + y + xy = - 1 \implies (x+1)(y+1) = 0$$

which would force either $$x = -1$$ or $$y = -1$$, a contradiction to $$x,y \in G$$.

Associativity
Next, we need to check associativity. We do this using the generic formula. We get:

$$(x * y) * z = (x + y + xy) + z + (x + y + xy)z = x + y + z + xy + yz + xz + xyz$$

and we also have:

$$x * (y * z) = x + (y + z + yz) + x(y + z + yz) = x + y + z + xy + yz + xz + xyz$$

Now, observe that $$*$$ is commutative (it is symmetric in $$x$$ and $$y$$). So it suffices to compute a one-sided identity element and verify the existence of one-sided inverses.

Identity element
First, we need to find the identity element. In other words, for any $$x \in G$$, we want:

$$x * e = x \iff x + e + xe = x \iff e (1 + x) = 0$$

Since $$x \ne -1$$, we get $$e = 0$$. Note that $$e \ne -1$$, so $$e \in G$$.

Inverse map
Finally, we need to compute the inverse map:

$$x * y = 0 \iff x + y + xy = 0 \iff y = -\frac{x}{1 + x}$$

This gives a formula for the inverse map. Note first that the formula makes sense, because $$x \ne -1$$, so $$1 + x \ne 0$$. Further, the output is not -1, because solving $$-1 = -\frac{x}{1 + x}$$ gives $$1 = 0$$, a contradiction. The inverse map is thus a well defined map from $$G$$ to $$G$$.

Thus, $$(G,*)$$ is a group with identity element $$0$$ and inverse map:

$$x \mapsto \frac{-x}{1 + x}$$

A group of symmetries
Here's another example. Suppose $$S$$ is a finite set of points in $$\mathbb{R}^3$$. Suppose $$G$$ is the set of all maps $$f: S \to S$$ such that for any $$x,y \in S$$, the distance between $$f(x)$$ and $$f(y)$$ equals the distance between $$x$$ and $$y$$. Define a binary operation in $$G$$ by composition:

$$(f * g)(x) = (f \circ g)(x) = f(g(x))$$

We want to show that $$(G,*)$$ is a group. Note that $$G$$ is realized as a set of functions under composition.


 * Closure of $$G$$ under $$*$$ follows from the transitivity of the relation of distances being equal.
 * Associativity follows from the fact that function composition is associative. Explicitly:

$$(f * (g * h))(x) = f((g * h)(x)) = f(g(h(x)))$$

and similarly:

$$((f * g) * h)(x) = (f * g)(h(x)) = f(g(h(x)))$$

Since this equality holds for every $$x \in S$$, we have:

$$f * (g * h) = (f * g) * h$$


 * The identity element is the identity map from $$S$$ to $$S$$. This clearly satisfies the condition for being an element of $$G$$.
 * To show that every map has an inverse, we first observe that any $$f:S \to S$$ that preserves distances must be injective. That's because if $$f(x) = f(y)$$, then the distance between $$f(x)$$ and $$f(y)$$ is zero, so the distance between $$x$$ and $$y$$ is zero, so $$x = y$$. Since $$S$$ is a finite set, $$f$$ must be bijective, so it has a unique inverse map. It is clear that this inverse map also preserves distances, so is in $$G$$.